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1 




AN 



ELEMENTARY TREATISE 



STATICS, 



By GASPARD MONGE. 



^itli a fiingrfl|iliital JMiu nf tti^ Sltttjinr. 



TRANSLATED FROM THE FRENCH, 



By woods baker, A.M. 



OF THE U:^aTED STATES COAST SURVEY. 



PHILADELPHIA: 

E. C. & J. BIDDLE, No. 6 S. FIFTH STREET. 

1851. 



Entered according to Act of Congress, in the year 1851, by 

E. C. & J. BIDDLE, 

in the Clerk's Office of the District Court of the United States, in and for the 

Eastern District of Pennsylvania. 



STEREOTYPED BY L. JOHNSON AND CO. 

PHILADELPHIA. 
PRINTED BY T. K. AND P. G. COLLINS. 






PREPACE 

TO THE 

AMERICAN EDITION. 



A WANT has long been felt in this country of a good 
elementary treatise on Theoretical Mechanics. The 
books on that subject, in the English language, are 
mostly voluminous, and either adapted to the compre- 
hension of those only who have mastered the various 
branches of analytical mathematics, or composed chiefly 
of practical and descriptive details. Hence, an accurate 
knowledge of the general laws, or grand fundamental 
truths of mechanics, so important to all men, in this 
eminently practical age and country, and especially to 
those who have some one of the useful arts as their 
daily occupation, has hitherto been attainable only by 
highly educated men. 

One of the consequences of the want of a familiar 

acquaintance with the mechanical laws, upon which all 

machines, of whatever possible kind, must depend, is 

the large number of failures of inventions occurring 

every year. The authors of such attempts generally 

3 



4 PREFACE. 

have mental ingenuity enough, but, unfortunately, they 
have not the knowledge necessary to render their con- 
trivances consistent with the laws of Nature, or adapted 
to attain the proposed ends by the best possible means. 
Machines, deficient in either of these essential matters, 
must sooner or later be discarded ; and their disap- 
pointed inventors have then to regret the loss of their 
money and time, which proper information would have 
prevented. 

To supply this urgent need, in part, the following 
little book upon Statics, or the science which treats of 
the equilibrium of forces applied to solid bodies, has 
been translated. It has long been known and highly 
admired by those who are familiar with the scientific 
literature of France ; but to persons who have little or 
no acquaintance with French authors on such subjects, 
it may be well to mention, that correct information, so 
well digested, precise and clear, can be obtained from 
the literature of no other nation or language. 

From the advertisement prefixed to the seventh 
French edition of this treatise, which has gone through 
eight editions in France alone, besides several that have 
been published at Brussels, the following extracts are 
made : 

'' The first edition of this work, which appeared in 
1786, was specially intended for young candidates 



PREFACE. g 

for the Navy; now, it is one of the standard books 
most generally followed. A correct and clear style, 
rigorous demonstrations, and well connected proposi- 
tions, have long caused it to be preferred for instruction 
in Statics. It is the first book in which there has been 
collected all that can be demonstrated in Statics syn- 
thetically. After having studied Euclid's Geometry, 
this work will be read without difficulty. Being mo- 
delled upon the method of the ancient geometers, it 
presents very clear ideas upon an abstract science, of 
which a great number of useful applications are made." 

'' A profound knowledge of Statics requires the aid 
of mathematical analysis, that is to say, of Algebra, and 
the DijQFerential Calculus ; but it is as important for be- 
ginners to study synthetical statics before analytical 
statics, as it is fit to precede the study of analytical 
geometry by that of elementary geometry." 

" The discussion of the elementary machines offers to 
M. Monge an occasion for showing the truth of a prin- 
ciple, which the author of the Mecanique Analytique 
(Lagrange) has rendered so productive, and which is 
known as the principle of virtual velocities.'' 

The treatise on Statics, of Monge, is a necessary 
introduction to the work of Poisson, which is a large 
and thorough analytical treatise, composed for the 
pupils of the Polytechnic School of Paris. 

1* 



6 PREFACE. 

The translator would here acknowledge his obligation 
to his friends, R, S. McCulloh, Esq., Professor of Na- 
tural Philosophy in the College of New Jersey, and 
J. B. Reynolds, Esq., Engineer and Lecturer on Me- 
chanical Philosophy before the Franklin Institute, Phila- 
delphia, for their kindness in giving him valuable advice 
and assistance in preparing this work for publication. 

Washington, Kov. 1850. 



BIOGRAPHICAL NOTICE 



GASPARD MONGE. 



It may be neitlier uninteresting nor out of place here^ to 
give a slight sketch of the life of the author of this work, 
who was one of the most distinguished of the eminent men of 
science in France at ihe close of the last century. It is 
selected from two papers,'^ written by his pupils^ MM. Brisson 
and Dupin, and presented by M. Delambre in his Analysis of 
the Labors of the Academy of Sciences diiring the year 1818. 

G-aspard Monge was born in 1746 His progress 

was so meritorious as to procure for him the Professorship of 
Physics in the College of Lyons, the year after he had com- 
pleted his studies in that institution Having 

gone to Beaune during vacation, he undertook to draw the 
plan of that city ; for which purpose he was obliged to make 
the necessary instruments. He presented his labor to the 
rulers of his native city, who compensated the young author 
as generously as the limited means of the public purse would 



* Notice Historique sur Gaspard Monge, by M. Brisson ; and Essoi Historique sur les 
Services et les Travaux Scientifiques de G. Monge, by M. Dupin. See the advertisement 
to the 7th edition of Geometrie Descriptive, par G. Monge, Paris, 1847. 

7 



8 BIOGRAPHICAL NOTICE. 

permit. A lieutenant-colonel of military engineers, who was 
tlien at Beaune, obtained for Monge the position of draughts- 
man and pupil in the school for engineers^ and conductors of 

works of fortification As he designed with rare 

ability, his manual talent only was regarded ; but he already 
felt his strength, and could not think, without indignation, of 
the exclusive esteem accorded to his mechanical skill. He 
said long afterwards : ^^ I was tempted a thousand times to 
tear my drawings to pieces through spite at the ado which 
was made about them, as though I had no ability for produc- 
ing any thing else.^^ The director of the school caused him 
to make some practical calculations of a particular case of de- 
filement, an operation serving to combine the relief and trace 
of a fortification, so that the defender may be sheltered from 
the fire of the assailant. Monge abandoned the mode pre- 
viously followed, and discovered the first geometric and general 
method that has been given for this important operation. . . 
.... By applying successively his mathematical talent to 
different questions of an analogous kind, and generalizing his 
means of conceiving and operating, he at length succeeded in 
forming a body of doctrine, which was his Descriptive Ge- 
ometry For more than twenty years he found it 

impossible to get the application of his geometry to draughts 
of carpentry taught to the corps at Mezieres. He was more 
fortunate, however, in the application to stone cutting; he 
carefully followed the methods employed in this art, and im- 
proved them by rendering them more simple by means of his 
geometry. 

His scientific labors secured him the nomination of Lecturer 
on Mathematics and Physics, to succeed Nollet and Bossut. 
He was afterwards appointed the regular professor, when, 
turning his attention to a scries of natural phenomena, he 
made numerous experiments upon electricity, explained the 
phenomena relating to capillarity, and was the founder of an 



BIOaRAPHICAL NOTICE. \) 

ingenious system of meteorology. He effected the decomposi- 
tion of water, and arrived at this great discovery without 
having had any knowledge of the researches made a short 
time before by Lavoisier, Laplace, and Cavendish. He was 
not satisfied with explaining the theories of science and their 
application to the students in the lecture rooms ; but liked to 
conduct his pupils wherever the phenomena of nature and the 
works of art could render these applications interesting and 
apparent to the senses. He infused into his pupils his own 
ardor and enthusiasm, and rendered delightful, observations 
and investigations, which, had they been viewed only abstractly 
in the narrow precincts of a lecture-room, would have appeared 
a wearisome study. 

In 1780, to attract Monge to Paris, he was associated with 
Bossut as Professor of the course of Hydrodynamics, estab- 
lished by Turgot. In order to reconcile the duties of the 
two places he had to fulfil^ he passed six months of the year 
at Mezieres, and six months at Paris. The same year he was 
elected a member of the Academy of Sciences; and, at the 
death of Bezout, in 1783, was chosen to replace that cele- 
brated examiner for the Navy. The Marquis de Castries fre- 
quently requested Monge to re-write the Elementary Course 
of Mathematics for the pupils of the Navy ; but he always 
declined. ^^ Bezout has left,^' said Monge, ^^a widow who has 
no other fortune than the writings of her husband, and I can- 
not think of snatching away the bread from the wife of a 
man who has rendered such important service to science and 
the country/^ The only elementary work which Monge pub- 
lished, was his Treatise on Statics ; and, with the exception 
of a few passages, where the evidence adduced might be some- 
what more rigorous, the Statics of Monge is a model of logic, 
clearness, and simplicity. 

At an epoch when the public distress called into the higher 
ranks all useful and courageous talent to the succor of the 



k 



10 BIOaRAPHICAL NOTICE. 

country^ whicli was threatened with an invasion, Monge was 
appointed Minister of the Navy. He did every thing to retain 
in France, men eminent for their merit or bravery ; and de- 
scended even to petition, to obtain the continuation of the 
services of Borda, in which he had the good fortune to suc- 
ceed. He was one of the most active men in scientific labors for 
the safety of the state. To him is due the construction of the 
new grinding apparatus in the powder works of Grenelle, and 
the boring machines on the boats of the Seine. He spent 
the daytime in giving instructions and directions in the work- 
shops, and the nights in preparing his Treatise on the Art of 
mahing Cannons; a work intended as a manual for the di- 
rectors of foundries, and for artizans. 

It was in his course of lectures at the Normal School that 
he delivered, for the first time, his Lessons on Descriptive 
Geometry, the secrets of which he had not been able sooner 
to reveal. Another institution, which preceded the Normal 
School in the order of conception, but which, being longer in 
maturing by its authors, followed it in the order of execution, 
realized in part the hopes that had been conceived in vain of 
founding the first Universal School opened in Prance. Monge, 
by uniting the result of long experience at Mezieres with his 
original and profound views, arranged the plan of studies, 
indicated their connection, and proposed practicable means of 
execution. Out of four hundred pupils, who had first entered 
the Polytechnic School, fifty of the best scholars were taken 
to form a preparatory school ; and it was Monge almost alone, 
who trained them. He remained the whole day amongst them, 
giving them by turns lessons in Geometry and Analysis ; ex- 
horting, encouraging, and inflaming them, by that ardor, be- 
nevolence, and impetuosity of genius, which caused him to 
exhibit to his pupils the truths of science with irresistible 
force and charm. In the evening, Monge commenced his 
labors anew : he wrote the pages of Analysis which were to 



BIOGRAPHICAL NOTICE. 11 

serve for the text of his next lessons ; and on the following 
day he was with his pupils at the first moment of their assem- 
bling. 

The amiability of Monge was neither the premeditation of 
the sage, nor even the effect of education : it was an unaffected 
gentleness of disposition, which he owed to his happy organi- 
zation. He was born to love and admire. He was excessive 
both in his admiration and his love ; hence he did not always, 
perhaps, remain within the limits where passionless and cold 

reason would have restrained him As he was the 

father of the pupils in the bosom of the school, he was like- 
wise the friend of the soldier in camp. 

While travelling through Italy, collecting the statues and 
paintings ceded to France by treaty, Monge had been struck 
by the strange contrast which the monuments of the Greeks 
presented to those of the Egyptians, transported to the 
banks of the Tiber by Augustus and his successors. The 
comparative character of^ the antique monuments became the 
subject of frequent conversation between the conqueror of 
Italy and the commissioner, who collected for his country 
the most beautiful of the fruits of victory. Monge con- 
ceived the idea of extending the domain of history beyond the 
fabulous ages of Greece ; of ascertaining, with the certainty 
of geometry, what the works of the ancient sages of the East 
were ; of recovering, by contemplating their monuments, that 
which had been — the processes of their arts, the customs of 
their public life, the order and majesty of their feasts and 
ceremonies. 

Charged by the General-in-chief, Napoleon, with bringing 
the treaty of Campo Formio to the Directory, he was, shortly 
afterwards, in the first rank of those who composed the Com- 
mission of Science and Art, which accompanied the Egyptian 
expedition. He was the first President of the Institute of 
Egypt, formed upon the model of the Institute of France. He 



12 BIOGRAPHICAL NOTICE. 

visited the Pyramids twice, saw the obelisk and the great walls 
of HeliopoliS; and studied the remains of antiquity scattered 
around Cairo and Alexandria. During a wearisome march in 
the interior of the desert, he discovered (as he supposed) the 
cause of that wonderful phenomenon known as the mirage. 
At the time of the revolt at Cairo, there were only a few de- 
tachments of troops in the city, and the palace of the Institute 
was guarded only by the scientific corps. It was proposed 
that they should make their way, sword in hand, to the head- 
quarters ; but Monge and Berthollet, considering that the palace 
contained books, manuscripts, plans, and antiquities, the fruits 
of the expedition, maintained that the preservation of this pre- 
cious deposit was the first duty of the scientific corps; and 
they decided to die, if necessary, in defending this treasure, 
rather than to desert it. 

Monge presided over the Commission of Science and Art of 
Egypt; and by his counsels, contributed greatly to the judi- 
cious conception of the plan, its arrangement, the proportion 
of the principal parts, and the means of improving the arts of 
execution. 

He had an inimitable manner of expounding the most ab- 
stract truths, and of rendering them clear by the language of 

action It was, however, only by combating nature 

that he became so excellent a professor : for he spoke with 
difficulty, indeed almost stammered; his utterance, causing 
him to drawl some syllables and utter others with too great 
rapidity. His physiognomy, habitually calm, presented an 
aspect of meditation ; but when he spoke, he appeared sud- 
denly as though a difi'erent man : a new fire instantly lighted 
up his eyes, his features became animated, and his figure in- 
spired 

Enfeebled by age, Monge became the victim of an imagina- 
tion, which, according as the times were adverse or propitious, 
carried him beyond well founded fears or hopes 



BIOGRAPHICAL NOTICE. 13 

The regulations of the service did not permit generous youth, 
at his funeral, to deposite the palm of gratitude and regret upon 
the tomb of their first benefactor; but, with the early dawn 
which followed the day of his obsequies, the pupils silently 
wended their way to his place of sepulture, and deposited there 
a branch of oak, to which they suspended a laurel crown. 
Twenty-three of the former pupils of the Polytechnic School, 
all residents of the city of Douai, united spontaneously, and 
decided to write in common to M. Berthollet, begging him to 
superintend the erection of a monument to be raised at the 
expense of the old pupils of the Polytechnic School, in honor 
of Graspard Monge. 



TABLE OP CONTENTS. 



PAGE 

Biographical Notice of Monge 7 

Definitions 15 

€HAP. I. — On the Composition and Decomposition of Forces 17 

II. — On Moments 56 

III. — On Centres of Gravity 87 

IV. — On the Equilibrium of Machines 118 

Art. I. — On the Equilibrium of Forces which 
act upon each other by means of 
Cords \ 120 

Art. II. — On the Equilibrium of the Lever 130 

Of Pulleys 141 

Of the Wheel and Axle 153 

Of Cog Wheels 166 

Of the Jack Screw 168 

Art. III. — On the Equilibrium of the Inclined 

Plane 169 

Of the Screw 185 

Of the Wedge 193 

NOTE, containing a Demonstration of the Parallelogram 

OF Forces, p.y M. Cauchy 203 

14 



AN 



ELEMENTARY TREATISE 



ON 



STATICS. 



DEFINITIONS. 

Every thing that is capable of affecting our senses is 
called a hody^ or material substance. 

Bodies are divided into solids and fluids, A body is 
solid when the molecules which compose it are cohesive, 
and cannot be displaced among each other without 
effort : the metals, stones, wood, &c., are of this num- 
ber. It is fluid when, on the contrary, all its molecules 
can be separated with the greatest ease ; such are water, 
air, &c. 

All bodies are moveable, that is, they may be trans- 
ported from one place to another. A body is said to be 
at rest, when all the parts vfhich compose it remain each 
in the same place ; and it is said to be in motion, when 
it changes place, or when the parts of which it is com- 
posed pass from one place to another. 

A body at rest cannot enter into motion, and when 
in motion, cannot change the manner in which it moves, 

15 



16 STATICS. 

without the action of some cause, to which has been 
given in general the name of force or power. 

There are to be considered in a force, Ist, its inten- 
sity^ that is to say, the effort which it makes to move 
the body, or the particle of the body to which it is ap- 
plied ; 2d, its direction^ or the straight line in which it 
tends to move the point of the body upon which it acts. 

When several forces are applied to the same body, 
two cases may happen : either these forces counter- 
balance and mutually destroy each other, when thej^ 
are said to be in equilibrium; or, by reason of the action 
of all these forces, the body enters into motion. 

Hence, the term Mechanics is given to that science 
whose object it is to find the effect which the applica- 
tion of determined forces must produce upon a body. 
This science is divided into two parts : the first con- 
siders the relations which the forces should have, in in- 
tensity and direction, so as to be in equilibrium, and is 
called Statics ; the second, to which the name Dynamics 
has been given, determines the manner in which the body 
moves, when these forces do not entirely destroy each 
other. 

Each of these parts is again divided into two others, 
according as the body, to which the forces are supposed 
to be applied, is solid or fluid. The part of Statics 
which treats of the equilibrium of forces applied to 
solid bodies, is named simply Statics, or Statics proper; 
and Hydrostatics is that which has for its object the 
equilibrium of forces applied to the different molecules 
of a fluid body. 

In this Treatise we will consider only the first of 
these two parts, that is. Statics proper. 



17 



G 



CHAPTER FIRST. 

OF THE COMPOSITION AND DECOMPOSITION OF FORCES. 

1. Whe7i a fo7^ce P, applied to a determined point c 
of a solid body ab^ draws or pushes this body in any 
direction CF, tve may consider this force as though it 
ivere immediately applied to any other "point D of the 
lody^ taTcen upon the direction of this force. 

For all the points of the body, which are 
in the straight line CF, can neither approach 
towards, nor remove from, each other; and 
none of them can move along this line with- 
i out causing all the others to move in the 
same manner as though the force were im- 
mediately applied to them. 

We are likeiuise permitted to consider the force P as 
though it were applied to any other point G, talcen beyond 
the body^ upon its direction^ ]jrovided this poiiit is in- 
valuably attached to the body. 

2. Hence it follows that if, upon the direction of the 
force P, there is found a fixed point D within the body, 
or an immovable obstacle beyond it, provided in the 
latter case the obstacle is invariably attached to the 
body, the force will be destroyed, and the body will re- 
main at rest; for this force may be regarded as imme- 
diately applied to the fixed point, and its efi*ect will be 
destroyed by the resistance of this point. 



{ 



18 STATICS. 

iig- 2. 3. Eeciprocally, if the force P, applied to the 

'\\}j body AB, is destroyed by the resistance of a 
single fixed pointy this point is found upon 
the direction of the force : for this point can 
destroy the effect of the force only by op- 
posing the motion of the point of application 
d c ; and it cannot prevent this motion, unless 
it is upon the straight line which the force tends to 
make the point of application traverse. 

4. A point cannot move in several directions at the 
same time. 

5. When several forces, differently directed, are ap- 
plied at the same time to the same point, either this 
point will remain at rest, or it will move in a single di- 
rection, and consequently in the same manner as though 
it were pushed or drawn by a single force along this di- 
rection and capable of the same effect. 

6. Thus, whatever may be the number and directions 
of the forces applied at the same time to the same 
point, there always exists a single force which can move 
it, or tends to move it in the same manner as all these 
forces together ; this single force is named the resultant 
of the former ; and they, with reference to the resultant, 
are named the component forces. 

The operation, by which we seek the resultant of 
several given component forces, is named the composi- 
tion of forces ; and that by which we find the compo- 
nents, when the resultant is known, is named the de- 
composition of forces. 

7. Two forces are equal^ when^ being applied to the 
same point and directly opposed^ they destroy each other 
and produce an equilibrium. 



COMPOSITION OF FOKCES. 19 

Reciprocally^ ivlun tivo forces are in equilibrium^ they 
are equal and directly opposed, 

8. For if several forces, differently directed, be ap- 
plied to the same point, it is necessary, in order to 
put them in equilibrium, or to destroy the effect of their 
resultant, that a single force, equal to this resultant, 
should be applied to this point and directly opposed to 
it, or that several forces be applied, the resultant of 
which is equal and directly opposed to the resultant of 
the former. 

9. Reciprocally, when several forces, differently di- 
rected and applied to the same point, are in equilibrium, 
their resultant is zero ; or what is the same thing, any 
one of these forces is equal and directly opposed to the 
resultant of all the others ; or, lastly, the resultant of 
any number of these forces is equal and directly opposed 
to the resultant of all the others. 

10. The resultant of two forces^ applied to the same 
pointy is in the plane determined hy the directions of these 
forces; and it is necessarily included in the angle 
formed hy these two forces. 

If the resultant were not in the plane of the two 
forces, there would be no reason why it should be above 
the plane rather than below : it cannot be at the same 
time in two different positions ; hence it is in the plane 
of the two forces ; moreover, it is included in the angle * 
of the lines along which these forces are directed; for 
there is no force tending to move the point into the 
space adjacent to this angle ; hence it will remain in 
the angle itself. 

11. A force is the multiple of another force when 



20 STATICS. 

it is formed hy the union of several forces equal to the 
latter. 

Thus, if we apply to the same point and in the same 
direction several forces equal to each other, and if we 
take any one of these forces as unity, the multiple force 
will be expressed by a number equal to that of the 
added forces. 

As it is always possible to compare numbers with 
lines, we may represent a force by a straight line taken 
upon its direction, and its multiple force by another 
straight line, a multiple of the former ; in Statics, fre- 
quent use is made of this geometrical construction. 

12. Three equal forces^ which are applied to the 
same pointj and which divide the circumference^ of which 
this point is the centre^ into three equal parts^ are ne- 
cessarily in equilibrium. 

The point to which these three forces are applied can 
move only in a single direction ; but the line along 
which it would move, could be placed in three or six 
manners perfectly similar with reference to the three 
forces ; now there is no reason that the motion of the 
point should take place in one direction rather than in 
another ; hence it will remain at rest. 

The application of these three equal forces to the 
same point proves, that there is evidently for this case 
a resultant of these forces, since any one of the three 
forces makes an equilibrium with the two others, and it 
is evident that the direction of one of these forces di- 
vides the angle formed by the two others, into two equal 
parts ; it is not less evident that the resultant of any 
two equal forces meeting in a pointy divides the angle 
formed hy these two forces into t^vo equal parts. 



COMPOSITION OF FORCES. 21 

13. If several forces are applied to the same point 
along the same line and act in the same direction^ it fol- 
lows^ from proposition No. 11, that their resultant is a 
single force^ equal to their sum, acting along the same 
line^ and in the same direction. 

Hence, to make an equilibrium of all these forces, it 
is necessary to apply to the same point and in the oppo- 
site direction, a force equal to their sum ; for this force 
will be equal and directly opposed to their resultant. 

14. From this it follows : 1st. If two unequal forces 
are applied to the same point in opposite directions, 
their resultant is in the direction of the greater, and is 
equal to their difference : for the greater of these two 
forces may be regarded as composed of two others 
having the same direction ; one of which is equal to the 
smaller force, and the other equal to the difference : 
now the first of these ^wo latter forces is destroyed by 
the smaller (7) ; then to move the point, there remains 
only the difference, which is in the direction of the 
greater. 

2d. If any number of forces are applied to the same 
point, some being placed in one direction, and the others 
directly opposed, after having taken the sum of all 
those acting in one of the two directions, and the sura 
of all those acting in the contrary direction, the re- 
sultant of all these forces is equal to the difference of 
the two sums (12), and is in the direction of the greater. 

Hence, to produce an equilibrium of all these forces, 
it is necessary to apply to the same point, and along 
the direction of the smaller of the two sums, a force 
equal to the difference of these sums ; for this force 
will be equal and directly opposed to their resultant. 



22 



STATICS. 



THEOREM. 



lb. If at the extremities of an inflexible straight 
line AB, two equal forces P, Q, he applied along the lines 
AP, BQ, parallel to each other^ and which act in the 
same direction : 

1st. The direction of the resultant B of these tivo 
forces is parallel to the straight lines ap, bq, and passes 
through the middle of ab ; 

2c?. This resultant is equal to the sum v+Q, of the 
two forces. 

DEMONSTiiATio:^r. — Let another in- 
flexible straight line DE be drawn 
perpendicular to the directions of the 
two forces P and Q, and invariably 
connected with the straight line ab : 
having prolonged the directions of 
the two forces, we may suppose that 
these two forces act (1) at the points 
D and E ; moreover, we may apply to the same points, 
the forces p^ p\ and q^ q\ equal to the forces P and Q, 
so that the three equal forces P, p^ p\ meeting at the 
point D, will divide the circumference, which has its cen- 
tre at the point D, into three equal parts, and so that 
the three forces Q, c[^ q^' equal to each other and to the 
first, will divide in like manner the circumference which 
has its centre at the point e, into three equal parts. 

We have seen (12) that the force P is equal and op- 
posed to the resultant of the couple p, p' ; that the 




COMPOSITION OF FOrtCES. 23 

force Q is equal and opposed to the resultant of the 
couple q^ q' ; hence the forces p and Q have a resultant 
equal and opposed to that of the system of the two 
couples 9, q' and p^ p' ; the two forces p' q\ applied to 
the point F of their direction, have for resultant a third 
force equal to each of the first two, and acting in the 
direction CF ; likewise the two forces 'p and 9, applied 
to the point a in the line of their direction, have for 
resultant a third force equal to each of them, and di- 
rected along the line go ; hence, the resultant of the 
four forces p^ p\ g, q' is a single force equal to two of 
the others, and acting along the line gcf, which divides 
the line de into two equal parts ; hence the resultant of 
the two equal forces p and Q is equal to the sum P + Q, 
and divides the line de, or the line ab, to which these 
forces are applied, into two equal parts. (See a second 
demonstration of this tTieorem, No. 19.) 



Corollary I. 

16. Hence, to bring the two forces p and Q into an 
equilibrium, it is necessary to apply to the middle K of 
the straight line ab a third force equal to their sum, 
which will act in a contrary direction, and which di- 
rection will be parallel to the two lines ap, bq ; for this 
third force will be equal and directly opposed to their 
resultant. 



Corollary II. 

17. If an inflexible straight line be divided into any 
number of equal parts, and we apply to all the points 



24 STATICS. 

of division, equal and parallel forces, the resultant of 
all these forces will pass through the middle of the line, 
in a direction parallel to that of the forces, and will be 
equal to their sum. 

For, all the partial resultants of these forces, con- 
sidered two and two, and taken at equal distances from 
the middle point of the line, will pass through this point 
(15), in the direction of the forces, and each of them 
will be equal to the sum of the two forces which com- 
pose it ; hence (13) the general resultant will also pass 
through the middle of the line, in the same direction, 
and will be equal to the sum of all the partial result- 
ants, that is to say, to the sum of* all the component 
forces. 



THEOREM. 

18. If at the extremities of an inflexible straight 
line, two unequal forces P and Q he applied, whose lines 
of direction AP, bq, are parallel to each other, and 
act in the same direction : 

1st, The resultant R of these two forces is equal to 
their sum, and its direction is parallel to that of these 
forces ; 

2d, The point of application c of the resultant di- 
vides the line ab into two parts reciprocally propor- 
tional to the two forces, so that we have 

p : Q : : BC : AC. 



COMPOSITION OF FORCES. 



25 




^^'^' Demonstration. — Suppose, in 

the first place, the two forces P and 
Q are commensurable ; divide the 
line AB into two parts ad, db, pro- 
portional to the forces p, Q ; laying 
off the line ad from A to E, and 
the line bd from B to F, the line ef will be double ab ; 
and since 

p : Q : : AD : DB, 
we shall have also, 

p : Q : : ED : df. 
Dividing the lines ad, db into as many parts as there 
are units in the forces P and Q^ and repeating this di- 
vision upon the lines ae, bf, the whole line ef will be 
divided into twice as many equal parts as there are 
units in the sum of the two forces P and Q ; now, the 
middle of each of these divisions may be considered as 
the points of application of forces equal to each other 

p_|_ Q 

and each equal to -^— (17), m being the number of 

units of the sum p + q; hence the resultant of all these 
forces will be also the resultant of the two forces p and 
Q ; but the resultant of equal forces, distributed on each 
side of the middle of equal divisions of the same line, 
is equal to the sum of these forces, passes through the 
middle of this line, and acts in the same direction as 
the components : hence the resultant of the two forces 
P and Q is equal to the sum P + Q, passes through the 
middle point c of the line EF, and acts in the direction 
of the forces P and Q ; moreover, from the construction 
of Figure 4, cf=Jef=ab; then subtracting the com- 
mon part CB, we have ac=bf=bd. For the same reason 

3 



26 



STATICS. 



ec=Jef=ab; hence cb=ae=ad ; hence the middle 
point c of EF is such that we have P : Q : : CB : CA ; 
hence this point of application divides the line ab into 
two parts reciprocally proportional to the forces. 

19. Let us suppose now that the two forces P and Q 
are incommensurable. Apply to the points A and b, 
and along the line ab, two equal and opposed forces B, s'; 
^^'^- the resultant of the two 

forces P and Q will be the 
same as the resultant of 
the four forces p, s, q, s' ; 
the couple P, S has for re- 
sultant a line Ai included in 
the angle sap ; the couple 
Q, s' has for resultant a line 
directed along bi included in the angle qbs' ; supposing 
these two resultants to be applied to the point of inter- 
section I of their directions, and drawing through the 
point I a line gh parallel to ab, they will each be de- 
composed into two forces along the directions IH and 
GH ; the forces in the direction gh being equal to s and 
s' and opposed, they will destroy each other ; the forces 
in the direction IR combine together and are equal to 
p+q; hence, whether the two parallel and unequal 
forces P and Q be commensurable or incommensurable 
their resultant is parallel to their direction and equal to 
their sum.* 




■^ This demonstration is also true when the two forces p and q 
are equal to each other ; consequently it is applicable to the theorem 
of No. 15. 



COMPOSITION OF FORCES. 



27 



D B.. 



"E 






-%• 4. 20. The resultant R of the two 

forces P and Q, supposed to be in- 
commensurable, passes through a 
point c in such a manner that we 
have 
^W rC p : Q : : CB :. CA. 

For if it passed through another point H, situated be- 
tween A and c, we should find a force q' applied in the 
direction Q, which would be such that the resultant of 
the commensurable forces P and q' would pass through 
a point K situated between the points c and H ; this 
force q' would be the fourth term of the proportion 

KB : KA : : p : q', 
the lines KB, ka, having for unit of measure a line 

hence the ra- 



smaller than CH ; but we have — > — 

' KA CA 



p — . p 

tio -7 is greater than the ratio -. Hence q' is smaller 

than Q : consequently, the resultant of the two forces 
p and Q will necessarily pass between the points K and 
q' ; hence it is absurd to suppose that it passes through 
the point H situated beyond the point K. It might be 
demonstrated in the same manner that it cannot pass 
through a point h' situated between c and B ; hence it 
passes necessarily through the point c. 

Corollary I. 



21. Hence, in order that equilibrium may ensue be- 
tween the two forces p, q, it is necessary to divide the 
line AB at the point c into two parts reciprocally pro- 
portional to these two forces, and apply to the point c 



28 STATICS. 

a third force, equal to the sum P+Q, parallel to them 
and acting in the opposite direction. 

Remaric, 

22. If the ratio of the forces p, q, and the length of 
the line ab, were given in numbers, and it were desired 
to find the distances of the point c from the points A, 
B, the proportion 

p : Q : : BC : AC 
could not be employed directly, because in this propor- 
tion we would know only the first two terms ; but it is 
easy to deduce from it the following, 

P+Q : Q : : BC+AC : AC, 
which, since bc+ac is equal to ab, becomes 

P+Q : Q : : ab ; AC, 
in which the first three terms are known. 

The distance BC can be found by the proportion 
P+Q : p : : AB : BC, 
which is likewise deduced from the first. 

COROLLABY 11. 

23. When a single force n is applied to a point c of 
an inflexible straight line, it may always be decomposed 
into two others p, Q, which, being applied to the two 
points A, B, given upon the same line, and being di- 
rected parallel to bc, produce the same effect ; and the 
intensity of the two forces is found by dividing the 
force R into two parts reciprocally proportional to the 
lines AC, cb, by means of the two following proportions : 

AB : BC : : B : p, 

AB : AC : : B : Q, 

in each of which the first three terms are known; for 



COMPOSITION OF FORCES. 29 

the resultant of the two forces p, Qj has the same in- 
tensity, is parallel to, and acts in the same direction as, 
the force R. 

Corollary III. 

24. Since Fig. 6 is the same in all respects as th^ 
preceding, if we apply to the point c of the line ab a 

^^' ' force s, equal and directly opposed to 

^/ the resultant of the two forces P, Q, in 

7? ^ ' ^ 

such manner that we have 

S=R=P + Q (19), 

the three forces p, Q, S, will be in equi- 
librium, and each of the two forces p, q, 
P" may be regarded as equal and directly 
opposed to the resultant of the two others. Hence the 
resultant of the two forces S, Q, which are parallel, and 
act in contrary directions, is a force p equal and directly 
opposed to the force p. Now, the force P is equal to 
the difference of the forces S, Q, and acts in the con- 
trary direction to the greater, s, of these two forces : 
hence, 1st, the resultant p or — p of the two forces S, Q, 
is equal to their difference S — Q, and acts in the direction 
of the greater and parallel to these two forces. 

Moreover, we have P+Q, or s : Q : : AB : AC (20). 

Hence, 2d, the distances of the point of application A 
of this resultant from the two points c, B, are recipro- 
cally proportional to the forces S, Q. 

Remark. 

25. If the ratios of the two forces S, Q, and the 
length of the line BC, were given in numbers, and 



30 STATICS. 

it were desired to find the distances of the point A 
from the points B, c, the preceding proportion could not 
be employed directly, because only the first two terms 
of this proportion would be known ; but it is easy to 
deduce the following from it : 

S— Q : Q : : ab— AC or BC : AC, 
in which the three first terms are known. 

The distance ab can be found by another proportion, 

S— Q : S : : ab— AC or BC : ab, 
which likewise is deduced from the first. 

Corollary IV. 

26. If the two forces S, Q, which are parallel, and 
act in contrary directions, be equal to each other, 1st, 
their resultant P, which is equal to S— Q (24), becomes 
zero ; 2d, in the proportion S— Q : Q : : BC : AC, the 
second term being infinitely great compared with the 
first which is zero, the fourth term is also infinitely great 
compared with the third. Hence the point of applica- 
tion A of the resultant p is at an infinite distance from 
the point c ; hence, to produce an equilibrium between 
the two forces S, Q, it would be necessary to apply to 
the inflexible straight line a zero force, whose line of di- 
rection should be at an infinite distance; which is not 
absurd, but which cannot be executed. 

We perceive, then, that it is impossible, by means of 
a single force, to produce an equilibrium between two 
equal and parallel forces, acting in contrary directions ; 
but it will be demonstrated (51), that, by means of two 
forces, an equilibrium can be produced between them in 
an infinite number of ways. 



COMPOSITION OF FORCES. 



31 



PROBLEM. 



27. Any numher of parallel forces p, Q, R, S, . . . . 
ivliich act in the same direction^ being applied to the 
points A, B, c, Bj . . . o of given position^ and connected 
together in an invariable manner^ to determine the re- 
sultant of all these forces. 

Solution. — Consider- 
ing first any two of these 
forces, such as P and Q, 
determine their resultant 
T (18) ; this resultant will 
be equal to P+Q; its di- 
rection will be parallel to 
that of the forces P, Q, 
and we wdll find its point 
of application E, by the 
following proportion (22) : 

p + Q : Q : : AB : AE. 
In place of the two forces p, q, substitute their re- 
sultant T ; then having drawn the line EC, determine 
the resultant v of the two forces T, R ; this resultant V 
will be also that of the three forces P, Q, R ; its intensity 
will be T+R or p + q+r; and find on EC its point of ap- 
plication F by the proportion 

T+R or P+Q+R : R : : EC : ef. 
In place of the three forces P, Q, R, let their re- 
sultant V be substituted, and having drawn the line fb, 
the resultant X of the two forces v, s, will be found ; 
this resultant x will be also that of the four forces p, Q, 
R, s; its magnitude will be v+s, or p + q+r+s, and 




32 STATICS. 

its point of application G upon fd will be found by the 
proportion 

v+s, or p+Q+R+S : S : : FD : Fa. 
By continuing in this manner, the position of the 
general resultant of all the forces will be found, what- 
ever their number may be ; and the intensity of this re- 
sultant will be equal to the sum of all these forces. 

Corollary I. 

28. Hence, by supposing the point G to be connected 
with the other points A, B, c, D, . . . . in an invariable 
manner, an equilibrium will be produced between all the 
other forces p, Q, R, S, . . . . by applying to the point G 
a force parallel to the first, which acts in the contrary 
direction, and which is equal to their sum 

P + Q + R + S, .... 

Corollary II. 

29. If among the parallel forces P, Q, R, s, . . . some 
should act in one direction, and others in the contrary 
direction, we could determine, first (27), the partial re- 
sultant of all which act- in one direction, and then the 
partial resultant of all which act in the contrary di- 
rection. Thus all the forces would be reduced to two 
others acting in opposite directions ; and by determining, 
by the process of No. 24, the resultant of these last 
two forces, supposed to be unequal, we should have the 
general resultant, and consequently the force which, 
being applied in the contrary direction, would produce 
an equilibrium amon^ all the proposed forces ; if these 



COMPOSITION OF FORCES. 33 

forces should be reduced to two parallel and equal 
forces, we have seen (26), that it is impossible to pro- 
duce an equilibrium between them. 

The general resultant being equal to the difference of 
the two partial resultants (24), and each of these being 
equal to the sum of those which compose it (27), it fol- 
lows that the general resultant is equal to the excess 
of the sum of the forces which act in one direction 
over the sum of those which act in the opposite di- 
rection. 

Corollary III. 

30. If the forces p, q, r, s, . . . without ceasing to 
be parallel and without changing in intensity, had 
another direction and became j9, g^, r, s, . . . the result- 
ant t of the first two w^uld also pass through the point 
E, and would be equal to the sum p+q. Likewise the 
resultant v of the three forces j9, q^ r, would pass through 
the point F, and would be equal to the sum p-\-q+r. 
So also the resultant x of the four forces ^, g, r, 5, 
would pass through the point a, and would be equal to 
the ^vim p + q+r+s^ and so on. Hence the general re- 
sultant of all the forces p^ q^ r, s, would pass through 
the same point as the resultant of the first forces 

P, Q, R, S, . , . . 

Hence it appears that, when the intensities and points 
of application remain the same, the resultant of these 
forces always passes through a certain identical point, 
W'hatever may be their direction ; and the intensity of 
this resultant is always equal to their sum. 



34 STATICS. 

The point througli wHcli the resultant of the parallel 
forces always passes, whatever may be their direction, 
is named the centre of parallel forces. 

It is easy to see that if the points of application A, 
B, c, D, . . . of the parallel forces P, Q, R, S, . • . are in 
the same plane, the centre of these forces is also in this 
plane ; for this plane contains the line ab, and conse- 
quently the point E of this line, which is the centre of 
the forces p, Q : it contains also the line EC, and conse- 
quently the centre F of the forces p, Q, r ; it contains 
the line fd, and consequently the centre G of the forces 
p, Q, R, S, and so on. 

It may be demonstrated in like manner that, if the 
points of application are upon the same straight line, 
the centre of the parallel forces is also upon this line. 

LEMMAS. 



31. jy^ a power v he applied to the circumference of 
^ff' ^' a circle movable about its centre A, and 

in a direction bp tangent to the circum- 
ference^ this force tends to turn the 
circle about its centre^ as though it were 
applied at any other point c, and in a 
direction CQ tangent to the same cir- 
cumference. 

II. 

A power p, applied along the direction of a line on 
which there is a fixed pointy is destroyed by the re- 
sistance of this point, (Nos. 2 and 3). 




COMPOSITION OF FORCES. 



85 



THEOREM. 

32. When the directions of two forces P, Q, are con- 
tained in the same plane^ and intersect in the same j 
point A, if we lay off in these directions the lines ab, 
AC, proportional to these forces^ so that we have 

p : Q : : AB : AC, 

and complete the parallelogram abcd, the resultant of 
these two forces will he in the direction of the diagonal 
AD of the parallelogram. 

■^- ^' Demonstration. — Suppose, 

for an instant, that the point 
D of the diagonal AD is an im- 
movable obstacle; from this 
point let fall the perpendicu- 
lars DE DF upon the directions 
of the two forces ; the trian- 
gles BED, CFD will be similar, 
because the angles at B and C 
being equal to A, will be equal 
to each other, and we shall have 

DC : DB : : DF : de. 

Now, we have by supposition 

p : Q : : AB : AC, or : : DC : db. 

Hence we shall have 




p : Q : : DF : de. 



36 STATICS. 

From the point D, as a centre, and with a radius DF, de- 
scribe the arc ra, terminating in the prolongation of ed 
at a ; then, regarding this arc and the line Ea as inflex- 
ible lines, and connected in an invariable manner at the 
point A, let us conceive the force P to be applied at the 
point E along its direction, and a force M, equal to the 
force Q, to be applied to the point a, in a direction 
parallel to AP, and consequently tangent to the arc 
ra. This being granted, since m=q and Dr=Da, we will 
have 

p : M : : Da : DE. 

Hence (18) the resultant of the two parallel forces 
p, M, will pass through the fixed point D, and will 
be destroyed by the resistance of this point ; hence 
these two forces will be in equilibrium around this 
point. 

Now the force Q, whose direction is tangent to the 
arc EG, and which we may regard as applied to the point 
r on its direction, tends to turn this arc in the same 
manner as the force M (31), and may be substituted for 
this latter force to counter-balance the force P : hence 
the two forces P, Q, will also be in equilibrium around 
the fixed point d ; hence their resultant will be de- 
stroyed by the resistance of this point, and conse- 
quently, (31) the direction of this resultant will pass 
through the point D. 

But the resultant of the two forces P, Q, should pass 
through the point of intersection A of their directions 
(4) ; hence this resultant will be in the direction of the 
diagonal ad. 



COMPOSITION OF FORCES. 



37 




33. Another Demonstration.* 
Suppose that the force Q acts at the 
point c on its direction, being inva- 
riably fixed at the point A, and 
that there is applied to the same 
point, c, and in the opposite direc- 
tions CMj cm', two forces M, m', 
each equal to Q, the effect of the 
two forces p and Q will be the same 
as that of the four forces P, Q, m, m', since the last two, 
M and m', destroy each other, being equal and opposed. 
Now, these four forces form two couples ; the one, Q and 
M intersect at the point c ; the other is composed of two 
parallel and unequal forces P, M^, applied to the line 
AC; the resultant of the two equal forces Q, M, is di- 
vided along the line CK, which divides (12) the angle 
MCQ into two equal parts ; the resultant hk of the two 
forces P, m', applied to the points A, c, of the line AC, 
passes through a point H of this line, so that we 
have (18) 

p : m', or Q : : HC : HA ; 
moreover, it is parallel to the direction ab of the force 
p. Hence the point k, the intersection of the two re- 
sultants CK, HK, is a point of the resultant of the four 
forces P, Q, M, m', and consequently of the first two P 
and Q. The point K is upon the direction of the di- 
agonal AD of the parallelogram constructed upon AB and 
AC, as sides; thus, by construction, the angle hck is 
equal to the angle kcd. Now, the angles kcd and ckh 



■^ This demonstratiou differs but little from that given by M. 
Poinsot, in the first edition of his Statique, in the year 1803. 



38 STATICS. 

are equal, being alternate internal angles ; hence, in the 
triangle CHK, the angles K and C are equal ; hence it 
follows, that the line kh has the same length as the line 
HC ; but we have the proportion : 

p : Q : : HC : HA ; 
hence we will have 

p : Q : : KH : ha, 
and since 

p : Q : : CD : AC, 

the ratio of the lines AC and CD is the same as that of 
the lines ah and hk ; hence the three points a, k, d, are 
in a straight line, and this line is the diagonal of the 
parallelogram constructed upon ab and AC as sides. 

Corollary I. 

34. If from any point D (Fig. 9), taken upon the 
direction AD of the resultant of the two forces p, q, the 
lines DB, DC be drawn parallel to the directions of these 
forces, a parallelogram abcd will be formed, whose sides 
AB, AC, will be proportional to the forces p, Q, that is to 
say, we will have : 

p : Q : : AB : AC, or : : DC : db. 

For if these sides were not proportional to the forces, 
their resultant would be in the direction of the diagonal 
of the parallelogram whose sides would be proportional 
to these forces (32), and not in the direction ad, which 
would be contrary to the supposition. 

Corollary IL 

85. If from any point D (Fig 9), taken upon the 
direction ad of the resultant of the two forces p, Q, 



COMPOSITION OF FORCES. 



39 



the perpendiculars, de, df, be drawn upon the directions 
of these two forces, these perpendiculars will be to each 
other reciprocally as the forces P, Q. 
For we have just seen (34) that 
p : Q : : DC : DB ; 
and the similar triangles dbe, dcf give 
DC : DB : : DF : DE ; 

hence 

p : Q : : DF : DE. 



THEOREM. 

36. When the directions of the tioo forces P, Q, are 
contained in the same plane^ and coincide in a point 
A, if the lines AB, AC, he laid off on these directions 
proportional to these forces^ so that 
p : Q : : AB : AC, 
and the parallelogram abdc he completed^ the resultant 
R of these tivo forces will he represented in intensity 
and direction hy the diagonal ad of the parallelogram ; 
that is to say^ we shall have : 

p : Q : R : : AB : AC : AD. 

Fig. 11. 

Demonstration. We have 
already seen (32) that the re- 
sultant of the two forces p, q, 
will be in the direction of the 
diagonal AD of the parallelo- 
gram; it is only necessary to 
show that its intensity will be 
represented by this diagonal. 




40 STATICS. 

Let there be applied to the point A, a force S, equal 
and directly opposed to the resultant R ; this force will 
be in the direction of the prolongation of the diagonal 
DAj and the three forces p, q, S, will be in equilibrium. 
Hence the force Q will also be equal and directly opposed 
to the resultant of the two other forces P, s ; and con- 
sequently this last resultant will be in the direction of 
the prolongation of the line CA. Let CA be laid off on 
its prolongation from A to H ; draw the line he, which 
will be parallel to ad, and consequently to the direction 
of the force S ; and through the point H draw HK parallel 
to the direction of the force P : the two forces p, S, will 
be to each other as the sides ab, ak, of the parallelo- 
gram ABHK (34), that is to say, 

p : s : : AB : AK or hb. 
Now, by reason of the parallelogram, ad=bh ; more- 
over, the two forces S and R are equal ; hence we have, 

p : B : : AB : AD. 
But by supposition 

p : Q : : AB : AC. 

Hence, by combining these last proportions, we will have 

p : Q : R : : AB : AC : ad. 



Corollary L 

37. If the two forces p, q, be applied to the point A, 
equilibrium will be produced by applying to the same 
point a third force in the direction AK, and proportional 
to the diagonal ad ; for this force will be equal and di- 
rectly opposed to the resultant of the two forces p, Q. 



COMPOSITION OF FORCES. 



41 



If the forces p, q be applied to other points of their 
directioHj equilibrium can be produced by applying to 
any point of the line ad, and in the direction DA, a force 
proportional to DA, provided the point of application of 
this last force is connected in an invariable manner 
with the points of application of the two forces P, Q. 



Fig. 11. 



Corollary IL 



38. A force R, of given in- 
tensity and direction, may be 
always decomposed into tvfo 
other forces p, Q, in the direc- 
tion of the given lines AP, AQ, 
provided these directions and 
that of the force R are included 
in the same plane and coincide 
in the same point A. 



For this purpose, we will represent the force R by a part 
AD of its direction ; then drawing through the point D 
the lines DC, db, parallel to the given directions ap, aq, 
we will form a parallelogram ABCD, whose sides ab, ac, 
will represent the required forces P, Q ; for (36) the re- 
sultant of these two forces will have the same intensity 
and the same direction as the force R. 

The intensities of the two forces P, Q, may be found 
by means of the proportions : 

AD : AB : : R : P. 

AD : AC : : R : Q. 




4* 



42 



STATICS. 



Fig, 11. 



/ \ 



H(- :^ 



Corollary III. 

39. In the triangle abd 
(Fig. 11), the sides AB, bd, ad, 
are proportional to the sines 
of the angles cad, dab, bac, 
formed by the directions of the 
forces p, Q, R ; hence it follows, 
that we shall have the propor- 
tion 



A A ^ A^ 

p : Q : R : : sin (q.r) : sin (p.r) : sin (p.q) 

A 
the notation (q.r), &c., being the angle of the two forces 

Q and r, &c. 

The same triangle abd gives 




hence 



ad^=ab^+bd^+2abxbd cos abd; 



A 



R^=p +Qil2PQ COS (p.q) 



PROBLEM. 

40. To determine the resultant of any numher of 
forces P, Q, R, S . . . . wliose directmis^ contained or 
not contained in the same jjlane^ meet in the same 
point A. 

Solution. Lay off upon the directions of all the 
forces, from the point A, the lines ab, AC, AD, ae . . . pro- 
portional to their intensities ; then, considering first any 
two of these forces, as P, Q, complete the parallelogram 
ABFC, whose diagonal af will represent in intensity and 
direction the partial resultant T of these two forces (36). 



COMPOSITION OF FORCES. 



43 




^'3' ^2- Instead of the forces 

p, Q, take their resultant 
T, and considering the 
two forces T, R, complete 
the parallelogram afgd, 
whose diagonal AG will 
represent in intensity and 
direction the resultant 
^ V of the two forces t, r, 
which will be that of the 
three forces p, q, r. 

In like manner, instead 
of the forces p, q, r, take 
their resultant V, and 
considering the two forces v, S, complete the parallelo- 
gram AGHE, whose diagonal ah will represent, in intensity 
and direction, the resultant x of the forces V, S, which 
will also be that of the four forces P, Q, R, S. 

By continuing in this way, the direction and intensity 
of the general resultant of all the forces p, Q, R, s . . . 
will be found, whatever may be their number. 

Corollary. 

41. If all the forces p, Q, R, S . . . . be applied to the 
point of coincidence A of their directions, in order to 
produce an equilibrium, jBnd first the intensity and di- 
rection of their resultant (40) ; then apply to the point 
A a force equal and directly opposed to it. But if the 
forces be applied to other points of their directions, 
connected together in an invariable manner, an equi- 
librium will be produced by applying 'to any invariable 
point in the direction of their resultant, a force equal 



44 



STATICS. 



and directly opposed to this resultant, provided the 
point of application of this force is also connected in an 
invariable manner with the points of application of the 
forces p, Qj Rj S . . . . 

PROBLEM. 

42. To determine the resultant of any number of 
forces^ wJiose directions included in the same plane do 
not meet in the same pointy whose points of application 
A, B, c, D . . . . are connected together in an invariable 
manner^ and whose intensities are represented by the 



parts Aa, b5j c^, dc?. 

Fig. 13. 



of their directions. 




/\ 



^K.\ 




Solution. Hav- 
ing prolonged the di- 
rections of any two 
of the forces, for in- 
stance P, Q, until they 
meet somewhere in a 
point E, lay off from 
E to F and from E to 
G the lines Aa, Bb, 
which will represent 
these forces ; and 
complete the paral- 
lelogram EF^G, whose 
diagonal Ee will re- 
present in intensity 
and direction the re- 
sultant T of the two 
forces P, Q (36). 



COMPOSITION OF FORCES. 45 

Instead of the forces P, Q, take their resultant T, and 
prolong its direction as well as that of the force R until 
they meet somewhere in a point H ; lay off the line Ee 
from H to Ij the line cc from H to K, and complete the 
parallelogram hiAk, whose diagonal rJi will represent in 
intensity and direction the resultant V of the two forces 
T, R, w^hich will be also that of the three forces P, Q, R. 

In like manner, instead of the three forces P, Q, R, 
take their resultant V, and prolong its direction, as well 
as that of the force S, until they meet in a point L ; 
then laying off from L to M and from L to N the lines 
hA, Ddy which represent the forces V and s, complete 
the parallelogram lmZn, whose diagonal lI will represent 
the resultant x of these two forces, which will also be 
that of the four forces P, Q, R, s. 

By thus continuing, the intensity and direction of the 
general resultant of all the proposed forces will be 
found, whatever may be their number. 

Corollary. 
43. Hence, when several forces, directed in the same 
plane, are applied to points connected together in an 
invariable manner, these forces always have one result- 
ant : thus it is possible to produce an equilibrium by 
means of a single force, except in the case where the 
direction of one of these forces being parallel to that 
of the resultant of all the others, the force and th^ re- 
sultant would always be equal to each other and would 
act in contrary directions ; for we have seen (26) that, 
in order to produce an equilibrium in that case, it would 
be necessary to apply a zero force in a line situated at 
an infinite distance ; which is impracticable. 



46 STATICS. 



THEOEEM. 

44. If three forces P, Q, R^ he rejyresented in intensi- 
ty and direction by the three sides AB, AC, ad, adjacent 
to the same angle of a parallelopipedon abfegd, so 
that 

p : Q : R : : AB : AC : AD, 

their resultant S will he represented in intensity and di- 
rection hy the diagonal AE of^ the p)arallelopipedon ad- 
jacent to the same angle^ and we shall have 

p : Q : B : S : : AB : AC : AD : AE. 

Demonstration. In the plane abcf, which contains 

the directions of the two forces p, q, draw" the diagonal 

-^'^* ^^' AE ; also draw the diagonal 

DE in the opposite face dheg: 

these tw^o diagonals will be 

&_/______^.^'''" equal and parallel; for the 

T^ ]/!--/- '"'•^"/ / ^^^ sides AD, EF of the paral- 

T. lelopipedon at the extremi- 
ties of which they terminate 







^ B p ^ ^^^ parallel and equal : hence 

AFED will be a parallelogram. This done, the two 
forces P, Q, being represented in intensit}^ and direction 
by the sides AB, AC, of the plane abfc, which is a paral- 
lelogram, their resultant T will be represented in mag- 
nitude and direction by the diagonal AF (36), and we 
shall have 

p : Q : T : : AB : AC : AF. 

Likewise the two forces T, R, being represented by the 
sides AF, AD, of the parallelogram afed, their resultant 



COMPOSITION OF FORCES. 47 

S, whicL. will be also that of tlie three forces P, Q, R, 
will be represented by the diagonal ae of the same par- 
allelogram, and we shall have 

T : E : S : : AF : AD : ae. 

Hence, combining the two proportions, we will obtain 
p : Q : R : S : : AB : AC : AD : ae. 

Now the diagonal AE is also the diagonal of the par- 
allelopipedon ; hence the resultant of the three forces 
p, Q, R will be represented in intensity and direction by 
the diagonal of the parallelopipedon. 

Corollary I. 

45. A force S of given intensity and direction can 
always be decomposed into three other forces P, Q, R, 
in the direction of the given lines ap, aq, ar, not in- 
cluded in the same plane, provided these three directions 
and that of the force "s coincide in the same point A. 

Thus, through the three directions, considered two 
and two, draw the three planes bag, cad, dab ; repre- 
sent the force S by a part ae of its direction; and 
through the point E draw three other planes EGDH, 
ehbf, efcg, respectively parallel to the first three. 
These six planes will be the faces of a parallelopipedon, 
of which AE will be the diagonal, and of which the sides 
ab, AC, AD, taken upon the three given directions, will 
represent the intensities of the three required forces P, 
Q, R ; for (44) the resultant of these three forces will 
have the same intensity and direction as the force S. 

Or else, draw through the point E three lines paral- 
lel to the directions ap, aq, ar ; and the parts ef, eh, 
EG, of these sides, included between the point E and the 
planes bag, cad, dab, will represent the intensities of 



48 STATICS. 

the required forces P, Q, R ; for these lineSj being three 
sides of the parallelopipedon, are respectively equal to 
the other sides ab, ac, ad, which are parallel to them. 

Corollary II. 

46. When the three forces P, Q, R, are perpendicular 
to each other, the resultant S is the diagonal of a rectan- 
gular parallelopipedon, whose three sides adjacent to 
the same summit of an angle are equal to the three 
forces p, Q, R ; the intensity of this resultant is, in this 
case, expressed by 

n/p^+qH^ 



Corollary III. 

47. Whatever may be the number of forces P, Q, R, 
S, . . . applied to the fixed points A, B, c, d, . . . we can 
always conceive the system of three right lines, perpen- 
dicular to each other, to be transferred to the points of 
application of the forces so as to occupy positions paral- 
lel to their former positions, and each of these forces to 
be decomposed into three others, in the directions of the 
three rectangular lines passing through the point of ap- 
plication ; then all the forces p, q, r, S, . . . will be 
decomposed into three systems of forces, so that all the 
forces of the same system will be reduced to a single 
force along the same direction ; hence all the forces P, 
Q, R, s, will have three resultants parallel to the three 
rectangular lines, fixed and determined in position with 
reference to these forces. (See No. 53). 



COMPOSITION OF FORCES. 49 



Corollary IV. 

48. Call S, S^5 s^' . . . . the forces whicli act upon a 
determined point ; and drawing through this point three 
lines fixed and perpendicular to each other, each of the 
forces S, b\ B^\ ... will be decomposed into three others 
p, q, r, in the direction of the rectangular lines. 

In like manner calling p\ q\ r\ the three component 
forces of the force s^ ; and p^\ q^\ r^\ the three com- 
ponent forces of the force s^', &c. ; the resultant of all 
the forces S, s', s'^ will be the diagonal of a rectangular 
parallelopipedon, whose three sides adjacent to the same 
angle will be, 

For the first, p+p'+p''+. . . . ; 

For the second, q+q^+q^^+. . . . ; 

For the third, r+r'+r''+ 

Hence the expression for this resultant will be 



THEOREM. 

49. Two forces in the direction of lines which do 
not meet cannot he reduced to a single force equivalent 
to them. 

Demonstration. Let p and q be the two forces 
whose directions do not meet. If a third force R causes 
an equilibrium to subsist between them, any two fixed 
points, one being taken upon the direction of this force 
R, and the other upon the direction of the force P, will 
necessarily destroy the force Q : now these two points 



60 STATICS. 

cannot be so taken that the line which unites them will 
not meet the force Q ; hence this force will not be de- 
stroyed ; hence it is absurd to suppose that the two forces 
p and Q can have a single resultant R. 



THEOREM. 

50. All the forces p, q, r, s, . . . applied to the 
points A, B, c, B, . . . joined together in an invariable 
manner^ may in general he reduced to two forces in the 
directions of lines which do not meet. 

• Demonstration. Having extended the lines in the 
directions of the forces p, Q, r, , . . until they cut a 
plane having a fixed and determined position with 
reference to these lines, we may consider the points of 
intersection as the points of application of the forces ; 
now each force may be decomposed into two, one situ- 
ated in the plane and the other perpendicular to this 
plane : all the forces directed in the plane will have one 
resultant; the forces perpendicular to the plane, and 
consequently parallel to each other, will have another 
resultant. In some particular cases, these two resultants 
meet, and all the proposed forces p, q, r, S, . . . will be 
reduced to a single one ; but in general they will not 
meet : hence we shall have two forces, one situated in a 
plane assumed arbitrarily, and the other perpendicular 
to this plane, which will produce an equilibrium with 
any number of forces p, Q, R, S, . . . . applied to the 
points A, B, c, D . . . . It is necessary to except from 
this general conclusion the particular case which we are 



COMPOSITION OF FORCES. 51 

about to examine, and which takes place when the forces 
situated in the plane and the forces perpendicular to 
the plane, are reduced to one or more couples of equal 
and parallel forces applied in opposite directions to the 
same right line. 

Corollary. 

51. When two forces act along the direction of lines 
which do not meet, there are an infinite number of sys- 
tems of two forces acting in the direction of other lines 
which do not meet, whose action is equivalent to that of 
the first two forces : in fact, any force may be decom- 
posed into two other forces, one perpendicular to a 
plane assumed at will, and the other situated in this 
plane ; hence any two forces are equivalent to two other 
forces, one being situated in the plane assumed arbi- 
trarily, and the other perpendicular to this plane. 

PKOBLEM. 

52. Two forces being given in the direction of lines 
which do not meet^ to find two others equivalent to theni^ 
one of which is in the direction of a line having 
a given position. 

Solution. Let p and q be the two given forces ; 
having drawn a plane perpendicular to the line of given 
position, decompose the forces p and Q into two others 
p' and q', one having its direction in this plane and the 
other perpendicular to the same plane ; decompose the 
force q', parallel to the given line, into two others q\ c/'^ 
parallel to q', of which the one q' will pass through the 



52 STATICS. 

given line, and the other q^' through a point of the force 
p' ; the two forces p', q^'^ meeting in the same point, 
will be reduced to a single force q ; hence the two forces 
P and Q will be transformed into two other equivalent 
forces g, c^^ the latter of which, g^, will pass through a 
given line. 

53. Examination of a particular case of the compo- 
sition of forces applied to given points A, B, c, d, . . . 
invariably joined together. 



Propositions. 

Having decomposed each of the forces p, q, r, s, . . . 
applied to the points A, B, c, D . . . into two others, one 
situated in a given plane K, and the other perpendicular 
to this plane, let S and T be the resultants of these two 
systems of forces ; it may happen that the first system of 
forces, instead of having a single force S for resultant, 
will only reduce to a couple of forces +s,-— s, equal, paral- 
lel, opposite, and applied to the same line ; in this case, 
the three forces, t, +5,— s, will be equivalent to two forces 
beyond the plane of the couple +§, —s: likewise, if 
instead of a single resultant T, the forces perpendicular 
to the plane K should reduce to a single couple +^, — ^, 
(designating by this expression two equal and parallel 
forces, opposed and applied to the same line), the three 
forces S,+^,— ^, will also be equivalent to two forces, as 
in the preceding case : finally, if all the forces acting in 
the plane K and perpendicularly to this plane, should be 
reduced to two couples +s,— s, and +^,— ^, these ^two 
couples would compose a single one. 



COMPOSITION OF FORCES. 53 

The three propositions "vvhich have just been an- 
nounced are included in the tYv^o following : 

1st. A force T and a couple +s, — s are equivalent to 
two forces in the directions of lines which do not meet ; 

2d. Two couples +s, —s and +^5 —-t are equivalent 
to a single couple of a like nature : such for example as 
+r, — r. 

Demonstration of these two Propositions. 

54. 1st Proposition. A force t and a couple +5,— s 
will combine into two forces : thus, the plane of the 
couple being prolonged intersects the force T in a point 
which may be regarded as the point of application of 
the force T ; drawing any line through this point and in 
the plane of the couple, and regarding this line as fixed 
with reference to the three forces t^+s^—s which are 
applied to it, decompose the force T into two other 
parallel ones t^ t\ which will have upon the fixed line 
the same points of application as the forces +§,-—5. 
The forces t^ s, meeting in the same point, will have a 
resultant; so also will the forces t\ —s: they will have 
a second resultant ; these two resultants evidently will 
be equivalent to the three forces t, +§,—<§. 

55. If the plane of the couple were parallel to the 
force T, it would be necessary to decompose the forces 

Fig. 13. {a.) of the couple+s and —s parallel to T. 

/ ij, Suppose this couple is applied to the 

/ _ ./ line AB (Fig. 13 a) perpendicular to 

H- -M this line ; drawing through their points 

/\^ / of application A and b, the lines am, 

/ / BN, parallel to the direction of the force 

T, decompose the force +s into two 



64 



STATICS. 



others in the directions am and AB, the force — s into 
two others bn and ba; the forces along the direction 
AB will be destroyed; the force along the direction am 
or BNj and the force T which is parallel to it, will com- 
bine to form a single force : hence, in this case, the three 
forces T5+S5— s, will reduce to two, one in the plane 
of the couple +§,—§, and the other parallel to this 
plane. 

56. 2d Proposition. Two couples +s,— sand +^,—^, 
situated in any planes w^hatever, will combine into a 
single couple+r,— r: thus, the planes of these couples 
being prolonged will meet in the direction of a line 
which may be considered as invariably joined to the 
points of application of the forces which compose the 
two couples. Let kl {Fig, 13 h) be this line, the in- 
tersection of the two planes lkmn, lkm'n^, one of which 
contains the couple +5,— s, applied to the line ab, the 
other the couple +tj—t^ applied to the line CD ; the 
directions of the forces +8^—8^+1^—1 intersect this 
line in the points a^ &, e^ d ; dividing the lines ab^ cd 
Fig. 13, (h,) into two equal parts, and marking the 
middle points /', /, we may transfer 
the couple +t,—t, applied to the line 
CD, parallel to itself, so that the points 
/, /' will coincide ; we will then have a 
new couple +t' ,--t\ It is necessary, in 
the first place, to demonstrate that this 
second couple, composed of forces equal 
and parallel to those of the first and 
applied to the line c'd' equal to CD, will be in equili- 
brium with it, and that in general we cannot change the 
condition of equilibrium of two couples, by transferring 




COMPOSITION OF FORCES. 55 

one of the couples parallel to itself in its plane. Now, 
this proposition is evident, for the point being the 
middle of the line c^d^ the two forces +t'^—t^ as well as 
the two forces —t\+t^ which act at equal distances 
from this point 0, are in equilibrium ; hence the second 
couple may be substituted for the first. Now, the force 
+ t' is decomposable into two other parallel forces pass- 
ing through the points a and h ; the force —t^ also may 
be decomposed into two others passing through the 
points b and a; and since c'a—d'l^ the component 
forces of these two forces ■\-t' and —%'^ will be equal, and 
will differ only in direction : hence, each of the three 
forces meeting in the point a will be equal to one of the 
three forces meeting in the point h ; hence, the result- 
ants of the two systems of forces applied to the points 
a and I will be equal and opposed ; hence it follows 
that the two couples +§,— s and -\-t^—t will be re- 
duced to a single one +r,— r. 

If the forces -\rt^—t were parallel to the line lk, the 
couple +5, — s (55) might be changed into another 
+/,—§', whose forces would be directed parallel to the 
line LK, and the four parallel forces +^,+/,— i^,— s' 
would be reduced to two equal and opposed forces 
+ (^+§0 and '-{t-\-%'Y 



^ See the theory of couples in the Statique of M. Poinsot, 6th 
edition, 1834. 



56 



CHAPTER SECOND. 



OF MOMENTS. 



57. Two kinds of moments are to be considered. The 
moment of a force referred to a point is the product of 
this force multiplied by the perpendicular let fall from 
the point upon the force ; the moment of a force referred 
to a plane^ is the product of this force multiplied by the 
distance of its point of application from the plane : this 
second kind of moment does not change^ even though 
the forces vary in direction ; they differ in this condi- 
tion from the moments of the first kind, which are inde- 
pendent of the points of application of the forces whose 
direction is constant. 

58. When the moments of several forces, referred to 
the same point, are considered, this point is named the 
centre of moments. 

59. Hence it follows, if the intensity of a force be 
known, and its moment referred to a centre or a plane, 
and if the plane be parallel to the force, we will obtain 
the distance of the centre, or the plane, from the di- 
rection of the force, by taking the quotient of the 
moment divided by the force ; if the moment and the 
distance be known, we shall obtain the intensity of the 
force by taking the quotient of the moment divided by 
the distance. 



MOMENTS. 



67 



COEOLLARY. 

60. When tlie two parallel forces p, q, act in the same 
direction, their moments, referred to any point c in the 
direction of their resultant, are equal. 



Fig. 15. 



--i) 



we shall have 



For if through the point c the 
line AB be drawn perpendicular to 
the direction of the two forces, and 
terminate in A and B, this line will 
be divided, at the point c, into two 
parts reciprocally proportional to 
the forces P, Q (18): that is to say, 



p : Q 



BC : AC. 



Hence, the product of the extremes being equal to 
the product of the means, we shall have 

PXAC=QXBC. 



THEOREM. 

61. If at the extremities A, B of an inflexible right 
line, two parallel forces P, Q be applied which act in the 
same direction, and if through the point of application 
c of their resultant, a right line de be drawn in any 
plane, upon which the perpendiculars AD, BE be dropped 
from A and b, lue shall have 



PXAD = QXBE. 



68 



STATICS. 



Fig. 15. 



\ 



JB,,f 



-fe" 



Demonstration. The right an- 
gled triangles ado, bec, which are 
similar, because the angles opposite 
the summit c are equal, give 
BC : AC : : BK : ad. 
"^Now we, have (18) 

p : Q : : BC : AC. 



Hence 



p : Q : : BE : AD ; 
and, the product of the extremes equalling that of the 



means, 



PXAD=QXBE. 



THEOREM. 



IX 



X 



y^ 



xio 



^i^' 16. 62. Two 2^(xraUel forces P, Q, 

H,y which act in the same direc- 

tioUj being applied to the points 
A, B of an inflexible right line^ 
and through a point r of this 
line, the line FH being drawn 
in any plane : 

1st. If the point F be taken 
upon the prolongation of ab, 
and if from the points A, B, 
and from the point of appli- 
cation c of the resultant^ the perpendiculars AG, bh, ci, 
be dropped upon fit, we shall have 

RXCI = QXBH + PXAG ; 



P^> 



MOMENTS. 



69 



Fig. 17. 



/ 



~7^ 



1^. 



2c?, If the 'point E he taken "be- 
tween A and B, 2^^ aAaZ? have 

RXCI=QXBH— PXAa. 

Demonstration. Through the 
point c, draw be parallel to eh, 
intersecting the perpendiculars ad, 
BH in the points b, e ; we have 



Da=ci==EH, besides (61), pxad=qxbe. 



y-\ 



X 



X 



t^ 



^^9- 16. Now, in the first case, the 

resultant R being equal to the 
sum of the two forces p, q 
(18), we shall have 

rxci=(q+p)xci, 
or 

=QXHE + PXaD. 

But we have GD=AD+Aa; 
hence 

rxci=qxhe+pxad+pxag; 
or placing the product QXBE, 
in place of pxad which is equal to it, we shall have : 

RXCI=QXHE + QXBE + PXAa; 

or finally 

RXCI=QXBH + PXAa. 

In the second case we have likewise 

RXCI=QXHE + PXGD. 

But we have gd=ad— ag; hence 

RXCI=QXHE + PXAD— PXAG ; 



60 



STATICS. 



or substituting for pxad its value qxbe, we shall have 

KXCI=QXHE + QXBE— PXAO ; 

and finally, 

EXCI==QXBH— PXAa. 



COEOLLARY I. 



Fig. 16. 



X 



3^r 






63. Hence it follows, 1st. 
When the two points A, b are 
on the same side of the line 
FH, the distance ci of the point 
c of the resultant from this 
line will be, since r=p+q, 

QXBH + PXAa 



Cl-= 



P+Q 



Fig. 17. 



y' 



•'<Et*« 



2d. When the points A, B are 
placed on different sides of the 
line EH, this distance will be 



ci=- 



QXBH— PXAa 



P + Q 
In this case the point c will be 
placed, with reference to the line 
GH, on the same side as the force Q, for which the pro 
duct QXBH is the greater. 



MOMENTS. 



61 



Corollary II, 



Fig. 16. 



>; 



c,- 



^^- 



and for Mg, 1 7, 



i^ey. 17. 



64. If the line FH is per- 
pendicular to AB, the lines 
AG, BH, CI will all three be in 
the direction of ab, and the 
proposition enunciated in the 
preceding theorem will still 
take place. In this case we 
shall have 



Then we shall have for 
Fig. 16, 
rxcf=qxbf+pxaf; 




RXCF=QXBr— PXAF. 

As to the distance CF from the 
point F to the point of applica- 
tion c of the resultant r=p+Q, 
it will be for JE^ig. 16, 

QXBF + PXAF 



CF=- 



CF = - 

and for Fig. 17, 

>XBF--PXAF 



P + Q 



P + Q 



THEOREM. 

65. Two parallel forces p, q {Figs. 18 and 19), 
which act in the same direction^ heing applied to the 
points A, B of an inflexible right line^ and having draivn 
a plane mn, through a point F of this line^ parallel to 
their directions : 



62 



STATICS. 



^^9' 18. 1§^. If the point F he 

taken upon the prolongation 
of AB, the sum of the mo- 
ments of the two forces P, Q, 
referred to the plane mn, 
will be equal to the moment 
of their resultant R : that 
is to say^ if from the points 
A, B, and from the point of 
- application c of the result- 
^ ant^ the peiyendiculars Aa, 

BH, cij he let fall upon the plane, we shall have 

RXCI=QXBH + PXAa. 




Ftff, 19. 



2d. If the point F he taken be- 
tween A and B, the difference of 
the moments of the force's p, q will 
he equal to the moment of their 
resultant : that is to say, we shall 
have 

RXCI=QXBH— PXAa. 



Demonstration. The three lines Aa, bh, ci, per- 
pendicular to the same plane mn, are parallel to each 
other; moreover they pass through the three points 
A, B, c, of the same line : hence they are in the same 
plane drawn through A, B ; hence their feet a, H, i, and 
the point r are in the same plane. But the four points 
E, Gj H, I, are also in the plane mn ; hence they are in 
the intersection of two diflferent planes, and conse- 
quently in a straight line. Now draw the line fgih : 




MOMENTS. 63 

it will intersect the lines Aa, bh, gi, at right angles ; for 
it will be in the plane MN" to which these lines are per- 
pendicular, and it will pass through their feet. Hence, 
by considering FGIH as the line fh {Figs 16 and 17) : 

1st. When the point F {Fig, 18) is upon the prolonga- 
tion of AB, we shall have (62) 

RXCI = QXBH + PXAG. 

2d. When the point F {Fig. 19) is between A and B, 
we shall have 

RXCI = Q + BH — PXAG. 
COEOLLARY. 

66. Hence, 1st. When the two forces p, q, {Fig. 18) 
are on the same side of the plane mn, the distance ci 
of the plane from the resultant will be equal to the sum 
of the moments of the forces referred to the plane, di- 
vided by the resultant R, or, what is the same (18), di- 
vided by the sum p+Q of the forces : that is to say, we 
shall have 

QXBH + PXAG 

P + Q 

2d. When the plane passes betw^een the directions of 

the two forces {Fig. 19), this distance will be equal to 

the difference of the moments divided by the sum of the 

forces : that is to say, we shall have 

QXBH — PXAa 
ci= ^ . 

P+Q 

In the latter case, the resultant will be situated on 
the same side of the plane mn as the force w^hose mo- 
ment is the greater. 



64 



STATICS. 



THEOREM. 



Fig. 20. 



67. If-, to any number of points A, B, c, D, . . . situ- 
ated or not in the same plane^ hut connected together in 
an invariable manner^ the parallel forces P, Q, R, S, . . . 
he applied^ which act in the same direction^ and which 
are all placed on the same side of any plane mn parallel 
to their directions^ the sum of the moments of all the 
forces referred to the plane mn will he equal to the mo- 
ment of their resultant. 

Demonstration. Draw the line ab, and let e be the 
point on this line through which the resultant T of the 

two forces P, Q pass- 
es ; draw the line 
EC, and let E be the 
point on this line 
through which the 
resultant v of the 
two forces T, R, 
passes, which will al- 
so be the resultant 
of the three forces 
p, Qj R ; draw ED, 
and let a be the 
point on this line through which the resultant x of the 
two forces v, S, passes, which will also be the resultant 
of the four forces p, q, r, s ; and so on. Finally, from 
the points A, B, c, D . . . . and the points E, E, a, . . . 
let fall upon the plane mn the perpendiculars Aa, bJ, c<?, 
D(^, . . . . 5 E6, f/, g^, . . . . 




MOMENTS. 65 

Then, the moment of the resultant t will be equal to 
the sum of the moments of its two components P, Q (65), 
and we shall have 

TXE^=PXAa + QXB5. 

In like manner, the moment of the force v will be 
equal to the sum of the moments of its two components 
T, R, and we shall have 

vxr/=TXEe+RXCe. 

Hence, substituting for TXEe its value, we get 

YXF/=PXAa + QXB5 + RXC(?. 

Likewise the moment of the force x will be equal to 
the sum of the moments of its two components V, S, 
which will give 

XXG^=VXF/+SXD6?. 

Hence, substituting for vxr/its value, we shall have 

XXGff=VXAa + QXBh + IiXCC + SX'Dd. 

And so on, whatever may be the number of forces. 
Hence the moment of any resultant is equal to the sum 
of the moments of all the components ; hence, &c. 



Corollary I. 

68. We have seen (27) that the intensity of the re- 
sultant X of the forces P, Q, R, S, . . . . is equal to the 
sum p+Q+R+s....of these forces: hence the dis- 
tance Gg of the direction of this resultant from the 
plane mn is equal to the sum of the moments of all the 



66 STATICS. 

forces P, Q, R, S, . . . divided by the sum of all these 
forces : that is to say, we shall have 

__PXAa + QXB5 + RXC^ + SXDcZ 
^^ P + Q + R + S • 



Corollary IL 

69. Hence, if an indefinite plane be drawn on the side 
on which the forces are placed, parallel to MN and at a 
distance ag from it : that is to say, equal to 

PXA(2 + QXBJ + RXC(? + SXD(i 
P+Q+R+S ' 

this plane will contain the direction of the resultant of 
all the forces p, Q, R, S . . . . ; for this plane will con- 
tain all the points w^hich, on this side, are distant from 
the plane MN by the quantity G^, consequently all the 
points of the direction of the resultant. 



Corollary III. 

70. If the forces P, Q, R, S, . . . he situated on each 
side of the plane MN, the moment of their result- 
ant^ referred to this plane^ tvill he equal to the excess 
of the sum of the moments of the forces tvhich are situ- 
ated on one side of the plane^ over the sum of the mo~ 
ments of the forces which are situated on the other side* 

Thus, let V be the partial resultant of all the forces 
P, Q . . . which are situated on one side of the plane, 
whatever their number may be, and E the point of ap- 



MOMENTS. 



67 



plication of this force. 
Fig, 21. 




In like manner, let x be the 
partial resultant of 
all the forces R, S, 
. . . which are situ- 
ated on the other 
side, and F the 
point of application 
of this force. By 
letting fall upon 
the plane the per- 
pendiculars Ka^ bS^ 
E6 . . . C{?5 Dc?5 f/, . . 
we have just seen 
(67) that we shall 
have 



and 



VXE6=PXAa + QXB5 . 



XXF/=RXCe + SXDcZ 



Now let Y be the resultant of the two forces v, x, and 
a its point of application ; this force will be the general 
resultant of all the forces P, Q, R, S, . . 

This done, the two forces v, x, being situated on each 
side of the plane MN, the moment of their resultant is 
equal to the difference of their moments (65) ; hence, 
letting fall the perpendicular Q^g upon the plane, we 
shall have 

' YXa^=VXE^--XXF/. 

Hence, by substituting the values of these last two mo- 
ments, we shall have 



68 STATICS. 

YXG^=PXAa + QXB6 . . . — {KXCC + SXJ)d . . .). 

Hence, &c. 



Corollary IV. 

71. Hence, in general, in whatever manner several 
parallel forces P, Q, R, s, . . . acting in the same direc- 
tion, may be situated with reference to a plane mn, 
parallel to their directions, the distance Gg of their re- 
sultant from this plane is equal to the excess of the sum 
of the moments of the forces situated on one side of the 
plane, over the sum of the moments of the forces situ- 
ated on the other side, divided by the sum of all the 
forces : that is to say, we shall have 

vXAa+QXBh . . . .—(nxcc+sx-Dd . . .) 
'^ P + Q + R + S ' 

and this resultant is placed, with reference to the plane 
MN, on the side on which the sum of the moments is the 
greater. 



Corollary V. 

72. Hence, if on the side of the plane mn, on Avhich 
the sum of the moments is the greater, we draw an in- 
definite plane parallel to it, which is distant by the 
quantity G^, or 

VXAa + QXBb . . . — (rXC(? + SXDcZ . . .) 

p+Q+R+S . . . ^ 



MOMENTS. 



69 



this plane will contain the direction of the resultant of 
all the forces p, q, r, S, . . . 



Corollary VL 



Fig. 22. 




YXG^=PXAa + QXB5 

Yxa^=pXAa+QXB5 



73. If the directions 
of the forces P. Q, R, S, . . . 
be all situated in the same 
plane, perpendicular to 
the plane mn, the lines 
Aa, bS, C(?5 D6? . . . . Q^g 
will fall upon the line kl, 
the intersection of the 
two planes ; and we shall 
then also have 

. + (RXC(?-rSXDi . . .), 
.— (rxcc?+sxdc? . . .)j 



according as the forces are on the same side or on the 
opposite sides of the line kl. Hence we shall have, in 
the first case. Fig. 22, 

_PXA(2 + QXb5 . . . + (RXC^ + SXD6? . . .) 
^^ P + Q + R + S... ^ 

and, in the second case. Fig. 23, 

PXA^ + QXb5 . . . — (RXC^ + SXD(i . . .)^ 



Gg~- 



P+Q+R+S 



70 



STATICS. 



Fig. 23. 




that is to say, when several 
parallel forces, situated in 
the same plane, act in the 
same direction, the distance 
of their resultant from any 
right line, traced in the 
same plane and parallel to 
their directionSjis in general 
equal to the excess of the 
sum of the moments of the 
forces situated on one side 
of the line, over the sum 
of the moments of the for- 
ces situated on the other 



side, divided by the sum of the forces. 



PROBLEM. 

74. An indefinite number of parallel forces being 
given, which act in the same direction and whose points 
of application may or may not be situated in the same 
plane, to determine, by means of moments, the direction 
of the resultant of all these forces. 

Solution. Having drawn at pleasure two different 
planes ABCD, BCIK, parallel to the directions of the 
forces, find the distance of the resultant from each of 
these planes separately (71) ; then draw a plane efgii 
parallel to abcd, at the distance from this latter plane 
that the resultant is from it, and situated on the side 
on which the sum of the moments referred to the plane 
ABCD is the greater ; then this plane efgh will contain 



MOMENTS. 



71 



the direction required (71). 
Fig. 24. 

B F 



X 



Likewise draw a plane 
LMNO parallel to bcik, 
at a distance from this 
latter plane equal to that 
of the resultant from it, 
and situated on the side 
on which the sum of the 
moments referred to the 
plane bcik, is the great- 
er; and this plane will 
also contain the direction 
required. Hence the direction of the resultant being 
both in the plane efgh, and in the plane lmno, it will 
be in the line of intersection PQ of these two planes. 



\7^ 



Corollary I. 

75. We have seen (30) that if several parallel forces 
change in direction, without changing either in intensity 
or their points of application, and without ceasing to be 
parallel to each other, their resultant always passes 
through the same definite point, which has been termed 
the centre of parallel forces; hence, for the parallel 
forces which we have just been considering, the centre 
is placed in the direction PQ of their resultant. 

To find this centre, draw at pleasure a third plane 
ABKR (I'lg. 24), and conceive that all the forces, with- 
out changing either their intensities or points of appli- 
cation, are directed parallel to each other and to the 
plane abkr ; find the distance of the resultant of these 
new forces from this plane (71). This done, if we draw 
a plane STVX parallel to abkr, and at a distance from 



72 



STATICS. 



this latter plane equal to that we shall have found, this 
plane will contain the new resultant, and consequently 
the centre of the forces. Hence the centre of the 
forces, being found both in the line PQ and in the plane 
STVX, it will be found in the point of intersection Y of 
the line and the plane ; or, which is the same thing, 
this centre will be found at the point of intersection Y 
of the three planes Eran, lmno, stvx. 



Corollary IL 

76. If the parallel forces p, q, r, S, . . . acting in the 

sante direction, be situated in the same plane ; in order 

Fig. 25. {a) ^ to find the position of their 

** ^-- — ^ — ^ ^ resultant, draw in this plane 

a line ln parallel to the direc- 
tions of the forces, and hav- 
ing let fall upon this line, from 
all the points of application 
A, B, c, D, . . . the perpen- 
diculars Aa, b5, C(?, D(^ . . ., 
lay off upon a line LM per- 
pendicular to LN, the line i^g'j 
equal to (73) 

PXAa + QXB6 + RXC(? + SXDC? . . . 




P + Q + R + S ... 

and the line ^'y, drawn through the point g' parallel to 
LN, will be the direction of the resultant. 

If all the forces were not on the same side of the line 
LN, it would be necessary to subtract the moments of 
the forces situated on the other side, instead of adding 
them (73). 



MOMENTS. 



73 



Fig, 25. (h) 



Corollary III. 

77. To find the centre of 




the parallel forces P, Q, R, S, 
. . . included in the same 
plane, conceive these forces, 
lyithout changing their intensi- 
ties and without ceasing to be 
applied to the same points 
A, B, c, D, . . . to be in direc- 
tions parallel to another line, 
as LM, upon which let fall the 
perpendiculars Ka\ b6', c^?', 

DC?', .... and the distance ^g' of this line from the new 

resultant will be (73) 

PXAa' + QXB5' + RXC(?' + SXDcZ' . . . 
P+Q+R+S ... 

Hence, if we lay off upon a line perpendicular to LM, 
the line i^g equal to this distance, and through the point 
g draw g^ parallel to lm, this line go. will be the direc- 
tion of the new resultant. Now, the centre of the forces 
is to be found both upon the direction of the first result- 
ant g'Y^ and upon that of the second ^a ; hence it will 
be at the point of intersection G of these two directions. 

If all the forces were not on the same side of the line 
LM, it would be necessary to subtract the moments of 
those which are situated on the other side, instead of 
adding them. 



74 STATICS. 



Corollary IV. 



78. If the points of application A, B, c, d, . . . {Fig. 
25) are in the same plane to which the directions of the 
parallel forces P, Q, R, S, . . . are oblique, the centre a 
of these forces will also be in this plane (30), and its 
position will be the same as though the directions of the 
forces were parallel to each other and situated in this 
plane. Thus, to find in this case the centre of the 
forces a, draw in the plane any two right lines ln, lm ; 
then suppose that the forces are in a direction parallel 
to LN, and find (77) the direction g'Y of their resultant 
on this supposition ; then suppose they are in a direction 
parallel to lm, and find the direction Qg of their re- 
sultant; the point of intersection a of the two lines 
^'y, ^G will be the centre of the forces required. 

The centre being found, if we draw through this 
point a line parallel to the real directions of the forces 
p, Q, R, S, . . . this line will be the direction of their 
resultant. 



Corollary V. 

79. Finally, if the points of application a', h\ c\ cZ', . . . 
{Fig. 25) be upon the same straight line lm oblique to 
the direction of the forces, the centre g' of these forces 
will be upon this line (30), and its position will be the 
same as though the directions of the forces were per- 
pendicular to LM. 



MOMENTS. 



75 



Hence (76) the distance ^'l of this centre from a 
given point l upon the line, will be equal to 



P + Q + R + S . . . 

If all the forces are not situated on the same side 
with reference to the point L, it will be necessary to 
subtract the moments of those which are situated on the 
other side, instead of adding them. 



LEMMA. 



Fig. 26. 



/^ 




80. When the directions of 
two forces P, Q, meet in a 'point 
A, the moments of these forces^ re- 
ferred to any point D in the line 
of direction of their resultant R, 
are equal. 

For we have seen (35) that if 
from the point D the perpendicu- 
lars DB, DC, be dropped upon the 
directions of the forces, prolonged 
if necessary, we shall have 



p : Q : : DC : DB. 

Hence, the product of the extremes being equal to 
that of the means, we shall have 



PXDB = QXDC. 



76 



STATICS. 



Fig, 27. 



Corollary. 

81. From this it follows that, if the directions of the 
two forces P, Q meet in a point A, the moment of any 
one, Q, of these forces, referred to 
a point D in the direction of the 
other, will be equal to the moment 
of their resultant R referred to the 
same point : that is to say, by let- 
ting fall from the point D, the per- 
pendiculars DB, DC upon the direc- 
tion of the force Q, and upon that 
of the resultant R, prolonged if ne- 
§ cessary, we shall have 

QXDB=RXDC. 

For, by applying to the point A a third force S, equal 
and directly opposed to the resultant R, the three forces 
P, Q, S, will be in equilibrium ; consequently the force p 
will be equal and directly opposed to the resultant of 
the two forces Q, s. Hence, the moments of the two 
forces Q, S, referred to the point D in the direction of 
their resultant, will be equal (80) ; hence we shall have 




QXDB=SXDC; 



or, smce S=R, 



QXDB=RXDC. 



MOMENTS. 



77 



THEOREM. 



Fig, 28. 82. Wlfien the directions 

of the two forces p, q, 
meet in the same point a, 
the moment of the result- 
\ ant R of these forces re- 

\ f erred to any point D, ta- 

\ hen in the plane of these 

^^\^ directions^ is equal to the 
^^ sum or difference of the 
- moments of the forces P, Q, 
referred to the same pointy ac- 
cording as the point D is with- 
out or within the angle paq, 
formed hy the directions of 
these forces : that is to say, if 
from the point D ive let fall 
'^ upon these directions, and 
upon that of the resultant, 
the perpendiculars DB, DC, de, 
we havcy in the first case, 




RXDE=QXDC + PXDB ; 



and, in the second case. 



RXDE = QXDC — PXDB. 
7* 



78 STATICS. 

Demonstration. DraM" the line ad, and decompose 
the force p into two others, p^ p'^ the first in the direc- 
tion AD, and the second in the direction of the force Q. For 
this purpose (38), represent the force P by the part af of 
its direction ; through the point F draw the lines Fa, fh, 
respectively parallel to aq and ad ; and the two com- 
ponents p, p' will be represented by the sides Aa, ah 
of the parallelogram agfh. 

The point D being upon the direction of the compo- 
nent p^ the moment of the other component y, referred 
to this point, is equal to the moment of their resultant 
Vy and we shall have (81) 

yXDC=PXDB. 

Moreover, taking the two forces ^, p' instead of the 
force P, the resultant R, of the two forces P, Q, is also 
the resultant of the three forces p, p\ Q. 

This being done, in the first case, the two forces Q 
and p^ {Fig. 28), which act along the same line of di- 
rection, are equivalent to a single force equal to their 
sum Q+y ; thus the force R may be regarded as the re- 
sultant of the two forces p and Q+y ; hence the moment 
of this resultant referred to the point D, in the direction 
of the first of these forces, is equal to the moment of 
the second (81) ; hence we shall have 

RXDE=(Q+y) DC, 

or 

RXDE = QXDC+yXDC. 



MOMENTS. 79 

Hence, by substituting for the moment p^XDC its 
value, we have 

RXDE = QXDC + PXDB. 

In the second case, the two forces Q, ;/ {^iff- 29), 
which are in the same line, and which act in contrary 
directions, are equivalent to a single force equal to their 
difference Q--p^ : now, the moment of this single force, 
referred to the point d in the direction of the force p^ 
is equal to the moment of the resultant R of these two 
forces (81) ; and we have 

RXDE = (Q--y)DC, 

or 

RXDE=QXDC— yXDC. 

Hence, substituting for the moment p'XDC its value, 
we have 

RXDE = QXDC~PXDB. 



HemarJc I. 

83. This theorem (82) of Statics is a consequence of 
the following geometrical proposition : 

If from any point D, taken in the 
plane of a parallelogram afml, we 
let fall the perpendiculars db, do, 
BE, upon the sides and the diagonal 
of this parallelogram, which meet 
in the same point a, the product 




D^^l^- 



80 



STATICS. 



of the diagonal am by its perpendicular de is equal to 
the sum of the products of the sides multiplied each by 
its perpendicular db and DC. Among the known demon- 
strations of this theorem, the following is one of the 
simplest : The triangles adf, adl, adm, having the same 
base AD, are to each other as their altitudes i/, L?, M.m ; 
but we have Mm=F/+LZ; for drawing lk parallel to 
AD, we have Mm=MK+Km=F/+LZ; hence the triangle 
ADM is equal to the sum of the triangles adf and adl : 
hence 

AMXDE=ArXDB+ALXDC. 



Remarh II. 



Fig, 28. 




84. If we suppose the line ad to be inflexible and the 
point D immoveable ; when this point is placed without 
the angle paq {Fig. 28), the two forces p, Q tend to turn 
the point A in the same direction around the point D ; 
and, on the contrary, when the point D is placed within 
the angle paq {Fig, 29), the two forces tend to turn the 
point A in opposite directions. 



MOMENTS. 81 

Hence, if two forces be directed in the same plane, 
the moment of their resultant, referred to any point 
taken in this plane, is equal to the sum or difference of 
their moments referred to the same point, according as 
these forces tend to turn their point of application 
around the centre of moments, either in the same or in 
opposite directions; and, in all cases, the resultant tends 
to turn its point of application in the same direction as 
that of the two forces whose moment is the greater. 



THEOREM. 

85. When the forces p, Q, R, S {Fig. 30), di- 
rected in the same plane^ are applied to the points 
a^ 5, (?, d^ connected together in an invariable manner^ 
tending to turn these points in the same direction around 
another point D taken in this plane^ the sum of the mo- 
ments of these forces^ referred to the point D, is egual to 
the moment of their resultant referred to the same point. 

Demonstration. Let v be the partial resultant of 
the two forces P, Q ; x that of the two forces v, R, and 
consequently of the three forces p, Q, r ; y that of the 
two forces x, s, and consequently of the four forces P, 
Q, R, S ; and so on. Then, from the point D let fall upon 
the directions of the forces and upon those of the par- 
tial resultants v, x, y, . , . the perpendiculars de, dp. 
Da, DH, . . . . Di, DK, DL, . . . This being done, the 
moment of the resultant v is equal to the sum of the 
moments of its components p, q, (82), w^hich gives 

VXDI=PXBE + QXDP. 



82 



STATICS. 

Fig, 80. 




H< 



/:-"" 






^'>% 



%i 



\ 






.-/.- 







In like manner the moment of the resultant X is equal 
to the sum of the moments of its components v, R, and 
we have 

XXBK=VXDI + RXDG ; 

or, substituting for the moment VXDI its value, 

XXDK=PXDE + QXDF + EXDG. 

Likewise, we have yxdl=xxdk+sxdh, or, substi- 
tuting the value of xxdk, 



YXDL = PXDE + QXDF + RXDa + SXDH; 



MOMENTS. 



83 



and so on, whatever may be the number of forces. 
Hence the moment of each resultant is equal to the sum 
of the moments of all its components. 



Corollary I. 

86. If the forces P, Q, R, S . . . . cZ(? not tend to turn 
their points of application in the same direction around 
the centre of the moments D, the moment of their re- 
sultant is equal to the excess of the sum of the moments 
of the forces which tend to turn in one direction^ over 
the sum of the moments of those which te7id to turn in 
the opposite direction. 



Fig. 31 . 



.^Tn 




A"! 






/-T-\. 







Thus, let V be the partial resultant of all the forces 
p, Q, . . . which tend to turn in one direction ; and let 
X be the partial resultant of all the forces R, s, . . . 
which tend to turn in the opposite direction ; from the 
point D let fall upon the directions of the forces and upon 
those of the two resultants v, x, the perpendiculars de, 
DF, . . . DG, DH, . . . Di, DK ; WO havc just Seen (85), 
that we have 



84 



STATICS. 



and 



VXDI = PXDE + QXDr , 



XXDK=IlXDa + SXDH 



Finallyj let Y be the resultant of the two forces v, x, 
and consequently that of all the forces p, q, r, s, . . . 

This being established, the moment of the resultant 
Y referred to the point d is equal to the difference of 
the moments of its components v, x, which tend to turn 
in opposite directions (84) : that is to say, by letting fall 
the perpendicular dl upon its direction, we have 

YXDL=VXDI — XXDK. 

Hence, by substituting the values of the two moments, 
we have 

YXDL=PXDE + QXDF . . . — (rXDG + SXDH . . .). 



Corollary II. 



Fig, 32. 



k 



% 



m 



%^ 



87. If the directions 
of the forces P, Q, R, S, . . . 
included in the same plane, 
are parallel to each other, 
the perpendiculars de, df, 
DG, DH, . . . DL let fall 



from the centre of mo- 
ments D upon these directions and upon that of the re- 
sultant Y, will be in the same straight line ; and the 



MOMENTS. 



86 



Fig, 33. 



preceding proportion will remain just the same, whether 
all the forces act in the same direction, as in Fig. 32, or 
whether they act, some in one direction and others 
in the opposite direction, as in Figs. 33 and 34. Now, 

the resultant Y of all these 
forces is equal to the excess of 
the sum of those which act in 
one direction, over the sum of 
those which act in the opposite 
direction (29) ; hence the dis- 
tance DL of the centre of mo- 
ments from the direction of 
the resultant, is equal to the 
quotient of the excess of the sum 
of the moments of the forces which 
tend to turn in one direction, over 
the sum of the moments of those 
which tend to turn in the opposite 
direction, divided by the excess of 
the forces which act in one direc- 
tion, over the sum of those which 
act in the contrary direction : thus 
we have 




Fig. 34. 




.^ 




DL 



DL 



DL 



f 



__PXDE + QXDF . . .— (rXDG + SXDH . . .) 



P + Q + R + S . . 


J 


___PXDE + QXDF . . .-— (rXDG + SXDH . 


••) 


P + Q . . . — (r + S . . .) 




_PXDE + RXDG . . . — (qXDF + SXDH . 


••) 



P+S . , .— (q + k . . .) 



{Fig. 32.) 
. {Fig. 33.) 
, {Fig. 34.) 



86 STATICS. 

In all cases, the resultant acts in the direction in 
■which the Sum of the forces is the greater ; and it is 
placed, with reference to the point b, on the side on 
which the sum of the moments is the greater. 






8T 



CHAPTER THIRD. 



ON CENTRES OF GRAVITY. 



88. The property by virtue of whicli bodies, left to 
themselves, fall towards the earth is named gravitation 
or gravity. 

All the molecules, of which bodies are composed, have 
gravitation, and they always have it ; for into whatever' 
number of parts a body is divided, each of these parts 
continually gravitates, and falls towards the earth when 
left to itself. 

89. The effort which a body makes to fall, when it is 
retained or supported by an obstacle which opposes its 
fall, is called the weight of the body ; this weight may 
be regarded as the effect of a force which is constantly 
applied to the body : thus we are accustomed to consider 
gravity as a force. 

Gravity is not a force rigorously constant for the same 
molecule ; it varies according to the different positions 
which this molecule has relatively to the sphere of the 
earth. 

1st. When the distance of the molecule from the cen- 
tre of the earth changes, its gravity decreases in the 
same ratio as the square of this distance increases ; be- 
sides, the earth not being perfectly spherical, and the 
lines drawn from its centre to the equator being greater 



88 STATICS. 

than those which terminate at the poles, the gravity at 
the surface of the earth is greater for the same mole- 
cule, when this molecule is placed at the poles, than 
when it is at the equator, because there the distance of 
the molecule from the centre of the earth is less. 

2d. The earth turns around its axis, and all the parts 
composing it perform their revolutions in the same time, 
that is to say, in about twenty-four hours. The parts 
of the surface near the equator describe greater circum- 
ferences of circles than those described by the parts 
near the poles ; their centrifugal force, which likewise 
is greater, destroys a greater part of the effect of gravi- 
tation, and is a new cause which renders this latter force ' 
less at the equator than it is at the poles. 

Thus, rigorously speaking, gravity is variable for 
the same molecule, 'when this molecule departs from or 
approaches the surface of the earth, and when it departs 
from or approaches the equator : but the distances of the 
positions in which we are accustomed, in Statics, to 
consider the same molecule, are so small with reference 
to the radius of the earth, that the effects of this varia- 
tion are absolutely insensible ; and we may regard 
gravity as a constant force for the same molecule, Avhat- 
ever its position may be. 

90. The straight line along which a molecule, aban- 
doned to itself, falls to the earth, and which is evidently 
the direction of gravitation, is named a vertical ; this 
line is everywhere perpendicular to the surface of the 
earth, or, more exactly, to the surface of undisturbed 
water. 

91. A plane is said to be horizontal when it is per- 
pendicular to a vertical. 



CENTRE OF aRAVITY. 89 

If the earth were perfectly spherical all the lines of 
direction of gravitation would meet in the same point, 
which would be the centre : but the earth not being a per- 
fect sphere, the lines of direction of gravitation for two 
different molecules may not be in the same plane ; and 
when they are in the same plane, they meet in the same 
point. 

However, the molecules in the same body, and those 
of the different bodies we are accustomed to consider in 
Statics, are so near each other compared with their 
distances from the centre of the earth, that the angle 
formed by the directions of gravitation for any two of 
them is not sensible, and we may regard all these direc- 
tions as parallel. 

92. We will regard, then, all the molecules of heavy 
bodies as constantly pushed or drawn towards the earth 
by forces constant for each of them ; we will suppose 
that these forces are parallel, and act in the same di- 
rection ; and, consequently, we will be able to apply to 
them all we have said of the composition, decomposition, 
and equilibrium of parallel forces. 

Now, when several parallel forces, acting in the same 
direction, are applied to points invariably connected to- 
gether, we have seen : 1st, that these forces have a re- 
sultant equal to their sum (27); 2d, that the direction 
of this resultant is that of the components ; 3d, that 
there exists a centre of forces through which this re- 
sultant always passes, even though the forces, without 
changing in intensity and without ceasing to be parallel^ 
should change in direction (30). 

Hence, 1st, the weights of all the molecules of a solid 
body have a resultant which constitutes the weight of 



90 STATICS. 

the body, and this resultant is equal to the sum of the 
weights of the molecules ; 2d, the direction of this re- 
sultant, or of the weight of the body, is always parallel 
to' that of gravitation, and consequently vertical; 3d, 
whatever may be the different positions given to this 
body, the directions of the resultants for all these posi- 
tions meet in the same point ; for by varying the position 
of the body, the intensity of the forces, which act upon 
the molecules, is not altered, and these forces, which 
only change in direction with reference to the bodies, 
do not cease to be parallel to each other. 

98. The point through which the direction of the 
weight of a body always passes, whatever may be its 
position, is named the centre of gravity. 

94. When several bodies are invariably connected to- 
gether, and we consider their assemblage as though they 
made but one and the same body, we ordinarily give to 
this assemblage the name of system, 

95. Every thing that has just been said of a single 
body may likewise be said of a system of several bodies : 
that is to say, the weight of the system is equal to the 
sum of the partial weights of the bodies which compose 
it ; that the direction of this weight is vertical, and that 
this direction, whatever may be the position of the sys- 
tem, always passes through the same definite point, 
which is the centre of gravity of the system. 



Corollary I. 

96. We may always regard the weight of a body, 
or system of several bodies, as a force directed verti- 



CENTRE OF GRAVITY. 91 

cally, and applied to the centre of gravity of the body 
or system : for this weight, which is the resultant of the 
partial weights of all the molecules which compose the 
body or system, may be considered as applied to any 
point of its direction, and consequently to the centre of 
gravity, which is always upon this direction, whatever 
may be otherwise the position of the body or the system. 



Corollary II. 

97. Hence, we will produce an equilibrium in the 
action which gravitation exerts upon all the molecules 
of a body or system of bodies, by applying to the centre 
of gravity of the body or system, a single force whose 
direction is vertical, equal to the total weight of the 
body or system, and which acts in a direction oppo- 
site to that of gravitation. 

Inversely, when a single force produces an equilibrium 
in the weight of all the molecules of a body or system 
of bodies, the direction of this force will be vertical, 
and it will pass through the centre of gravity of the 
body or system. 

Thus, when a body ab, suspended by a 
thread ed from a fixed point D, is in equi- 
librium, and the action of gravitation is con- 
sequently destroyed by the resistance of the 
thread, the direction of this thread will be 
vertical, and its prolongation will pass 
through the centre of gravity c of the body. 





^ STATICS. 

Corollary III. 

98. From this, vfe deduce a simple man- 
ner of finding, by experiment, the cen- 
tre of gravity of a body of any figure. 
Thus, if we suspend the same body by a 
thread, successively by the two different 
points E, E^, and conceive the two directions 
of the thread to be prolonged into the interior of the 
body, the point c, in which these two directions inter- 
sect, will be the centre of gravity required. 

Remark. 

99. Since the partial weights of the bodies which 
compose a system, may be considered as parallel forces, 
applied to the partial centres of gravity of these bodies, 
it follows, that when we know the weight of these bodies, 
and the positions of their partial centres of gravity, we 
can find the position of the centre of gravity, by the 
processes which have been given to find the centres of 
parallel forces, either by means of the principle of the 
composition of parallel forces, as in No. 28, or by em- 
ploying the consideration of moments, as in Nos. 75, 
77, 78, and 79 ; we will soon have occasion to give ex- 
amples. 

To find the centre of gravity of a body of any figure 
whatever, conceive the body to be divided into a certain 
number of parts, so that we may know the weight of 
each of them, and the position of its partial centre of 
gravity ; then, by finding the centre of gravity of the 
system of all these parts, we will have the required cen- 
tre of gravity of the body. 



CENTRE OF GRAVITY* 93 

But when the parts of the body are of the same na- 
ture in all its extent, and when the figure of the body 
is not very complicated, we may often find its centre of 
gravity by simpler considerations, and which we are 
about to employ in order to arrive at results which are 
used very frequently. 




LEMMA. 

100. When a body is considered 
to he composed of parts A, a', b, b', 
c, c', . . . which^ taken two and two^ 
A are equal to each other ^ and so placed 
thqt the middle of the lines AA^, 
bb', cc', which Join the centres of 
gravity of the homologous parts ^ co- 
incide in the same point D, this pointy which is the 
centre of figure of the body, is also its centre of gravity. 
For the point D is the centre of gravity of each par- 
tial system of two homologous parts ; hence, it is also 
the centre of gravity of their general system. 



Corollary. 

101. By considering lines, surfaces and solids, as 
composed of parts uniformly heavy, it is evident : 1st, 
that the centre of gravity of a right line is at the mid- 
dle of its length ; 



94 



STATICS. 




2d. The centre of gravity of the area, 
and that of the contour of a parallelo- 
gram ABCD, are in its centre of figure : 
that is to say, at the point of intersec- 
tion E, of its two diag-onals AC, bd ; 

3d. The centre of gravity of the area of a circle, and 
that of its whole circumference, are at the centre of the 
circle ; 

4th. The centre of gravity of the whole surface of a 
parallelopipedon, and that of its solidity, are in its cen- 
tre of figure : that is to say, in the intersection of any 
two of its four diagonals, or in the middle of one of 
them; 

5th. The centre of gravity of the convex surface of 
a right or oblique cylinder, and that of its solidity, are 
in the middle of the length of its axis ; 

6th. The centre of gravity of the surface of a 
sphere, and that of its solidity, are at the centre of the 
sphere. 



PROBLEM. 



102. To find the centre of gravity of the area of any 
rectilinear triangle abc. 

Fig. 38. Solution. Having drawn through 

the summit A of one of the angles, 
and through the middle D of the op- 
posite side, the line ad, if we con- 
ceive the area of the triangle to be 
divided into an infinite number of 
elements by lines parallel to BC, the 




CENTEE OF GRAVITY. 



95 



centre of gravity of each of these elements will be in 
its middle (101), and consequently upon the line ad; 
hence the centre of gravity of their system, which will 
be that of the area of the triangle, will be upon this 
same line (30). For the same reason, if from the sum- 
mit B of another angle, and through the middle E of 
the opposite side, we draw a line de, this second line 
will contain the centre of gravity : hence this centre 
will be found both upon the line ad, and upon the line 
BE ; hence it will be found at the point of intersection 
F of these two lines. 



Corollary I. 

^^9' 38. 103. If from the summit A of one 

of the angles of the triangle ABC, 
and through the middle D of the oj^ 
posite side, we draiv a line ad, and 
divide this line iyito three equal parts^ 
the centre of gravity E of the area 
of the triangle tvill he upon this line, 
at the distance of two-thirds from the summit of the 
angles, or one-third from the opposite side. 

For, if we draw the line de, this line will be parallel 
to AB, because the sides bc, ac, are cut proportionally 
in D and e ; and the triangles abf, def will be similar, 
because their corresponding angles will be equal ; hence 
we shall have 




AF : FD : : AB : de. 
But the similar triangles abc, edc, give 



96 STATICS. 

AB : DE : : EC : DC, or : : 2 : 1 (102). 
Hence we shall have 

AF : FD : : 2 : 1, or af=2fd, 



hence, 



fd=Jad, and af=|ad. 



Corollary IL 

104. If in the plane of a 
rectilinear triangle ABC we draw 
any line Gi, the perpendicular 
let fall from the centre of gravi- 
ty F of the area of the triangle 
upon Gl, will be equal to one- 
third of the 8U7n of the perpen- 
diculars AG+CH+BI let fall 
'from the summits of the angles 
upon the same line. 

Thus, through the summit A of one of the angles 
draw the line am parallel to gi, which will cut in K, L, 
M, the perpendiculars let fall from the other points ; 
through the point A, and through the centre of gravity 
F, draw the line af, whose prolongation will bisect the 
opposite side at the point D ; finally, through the point 
D, draw DN perpendicular to am : this being done, we 
shall have 




I>N= 



CK + BM 



CENTRE OF GRAVITY. 97 

and the similar triangles afl, adn will give 

FL : DN : : AF : AD, or : : 2 : 3 (103). 
Hence, 

CK + BM 
FL=fDN= g . 

But the lines AG, kh, lo, mi, being equal to each 
other, we shall have 



L0 = 



AG + KH + MI 



Hence, by adding this equation to the preceding, we 
shall have 



FL + L0 = 

that is to say, 



AG + CK + KH + BM + MI 



F0 = 



AG + CH + BI 



From this we deduce 
another manner of finding 
the centre of gravity of 
^' the area of a rectilinear 
triangle ABC. Having 
drawn at pleasure, in the 
plane of the triangle, two 
_ lines Gl, GP, and having 
found the distances FO, 
FR, of the centre of gravity from each of these lines, 




98 



STATICS. 



if we draw the line RV parallel to Gl and at the distance 
PO, this line will contain the required centre of gravity; 
in like manner if we draw xo parallel to PG and at the 
distance m, this second line will contain the centre of 
gravity ; hence this centre will be found upon the two 
lines RV, xo ; hence it will be at their point of inter- 
section P. 



PEOBLEM. 

105. To find the centre of gravity of the area of a 
rectilinear polygon abcde of any nurriber of sides. 



First Solution, 
of parallel forces. 

Fig, 41. 



hy the process of the composition 



Divide the area of the polygon 
into triangles by the diagonals 
AC, AD, . . . drawn from the sum- 
mit of the same angle A, and de- 
termine (102, or 103, or 104), 
the partial centres of gravity F, 
G, H of the areas of these trian- 
gles ; then considering these tri- 
angles as weights proportional 
to their areas and applied to their centres of gravity, 
join the centres of gravity of the first two triangles abc, 
CAD by a line fg, and find upon this line the centre of 
gravity I of the system of the two triangles, or of the 
quadrilateral abcd, by dividing the line fg into two 
parts reciprocally proportional to the areas of the two 




CENTRE OF GRAVITY. 



99 



triangles (18,) which may be done by the following pro- 
portion (25) : 

quadrilateral abcd : triangle cad : : fg : Fl. 

Through the point I, and through the centre of gravity 
H of the next triangle, draw the line hi, upon which 
find the centre of gravity K of the system of the first 
three triangles, by dividing this line into two parts re- 
ciprocally proportional to the areas of the quadrilateral 
ABCD and of the triangle dae, which may be done by 
the following proportion : 

pentagon abcd : triangle dae : : ih : ik. 

By thus continuing, whatever may be the number of 
triangles, we will find the centre of gravity of their sys- 
tem, and this centre will be that of the area of the pro- 
posed polygon. 

Second Solution, tahen from the consideration of 
moments. 



m\ 



F'^9' 42. Having divided the area 

of the polygon, as in the 
preceding solution, and de- 
termined the partial centres 
of gravity F, G, H, . . of all 
the triangles, draw at plea- 
sure in the plane of the poly- 
gon two lines LM, ln, upon 
which let fall perpendicu- 
lars from all the centres of gravity F, G, H, . . ; con- 
sider these lines as the intersection of two planes paral- 




hy?i' 



100 STATICS. 

lei to the direction of gravitation. This being done, the 
distance of the centre of gravity of the polygon, or of 
the system of all the triangles, from each of the lines 
LM, LN, will be equal to the sum of the moments of the 
triangles referred to each plane, divided by the sum of 
their areas (77) : thus the distance of this centre from 
the line lm will be 

ABCXF/zhCADXG^dzDAEXHA 
ABODE ' 

and its distance from the line ln will be 

ABC X F/=bCAD X G^'dzDAE X hA' 
ABODE 

Hence, by drawing a line parallel to lm, and at a 
distance equal to the first of these two distances, this 
line will contain the centre of gravity of the polygon ; 
likewise if we draw a line parallel to ln, at a distance 
equal to the second of these distances, this line will con- 
tain the centre of gravity ; hence the intersection of 
these two lines will be the required centre of gravity. 

106. If the centres of gravity r, a, h, of the trian- 
gles which compose the area of the polygon, were not 
all placed on the same side with reference to each of the 
lines LM, LN, in order to find the distance of the centre 
of gravity K of the polygon from each of these lines, 
it would be necessary to subtract the moments of the 
triangles, whose centres of gravity are situated on the 
other side of this line, instead of adding them (77). 



CENTRE OF GRAVITY. 



101 



PHOBLEM. 

107. To find the centre of gravity of the contour of a 
polygon abode, of any number of sides. 

First Solution, ly the process of the composition of 
parallel forces. 

^^9' 43. Bisect each of the sides 

of the polygon at the 
points F, a, h, i, k, which 
will be the partial cen- 
tres of gravity of these 
sides (101). Then, con- 
sidering all the sides as 
weights proportional to 
J" ^' y' ^' ^ their lengths, find the 

centre of gravity of the system of any two of them, 
as AB, BC, by joining their centres of gravity by the line 
Fa, and dividing this line into two parts reciprocally 
proportional to these sides, which may be done by the 
following proportion (22) : 

AB+BC : BC : : Fa : FO. 

The point being found, draw through this point, and 
through the middle H of the next side, the line oh, upon 
which find the centre of gravity P of the system of the 
three sides, by dividing this line into two parts recipro- 
cally proportional to the side CD and the sum of the first 
two, AB, BC ; which may be done by the proportion, 




AB-f BC+CD : CD : : OH : op. 



102 i STATICS. 

In like manner, drawing the line pi, find the centre 
of gravity Q of the system of the four sides AB, BC, CD, 
DE by the proportion, 

AB + BC + CD + DE : DE : : PI : pq. 

By thus continuing, whatever may be the number of 
sides of the polygon, find the centre of gravity of their 
system, and this centre will be that of the contour of the 
polygon. 

Second Solution, tahen from the consideration of 
moments. 

Having bisected each side of the polygon, draw at 
pleasure the two lines lm, ln, upon each of which let 
fall perpendiculars from the middle of all the sides. 
This being done, the distance of the centre of gravity 
R of the system of all the sides referred to each of the 
planes, whose lines of intersection are lm, ln, will be 
equal to the sum of the moments of the sides referred 
to this plane, divided by the sum of the sides (77); 
thus, the distance of this centre from the line lm will be 

ABXYf+BCXag + CBXllh + DEXli+T^AXKk 

ab+bc+cd+de+ea ^ 

and its distance from the line ln will be 

ABXF/ + BCXG/ + CDXHA' + DEXr/' + EAXK>fc 
AB + BC + CD + DE + EA * 

Hence, drawing a line parallel to lm, at a distance 
equal to the first of these distances, then another line 
parallel to ln, at a distance equal to the second of these 



CENTRE OF GRAVITY. 



103 



distances, the point of intersection of these two lines 
will be the centre of gravity R of the contour of the 
polygon. 

RemarTc, 

108. If the middle points F, a, H, i, K {Fig. 43), 
of the sides of the polygon, were placed on opposite 
sides of the lines lm, ln ; in order to find the distance 
of the centre of gravity R from each of these lines, it 
would be necessary to subtract the moments of the sides 
whose middle points were situated on the contrary side, 
instead of adding them (77). 



PROBLEM. 

109. To find the centre of gravity of the solidity of 
any triangular pyramid abcd. 

Solution. Determine the centre 
of gravity F of the area of one of 
the faces bcd of the pyramid (103), 
by drawing through the summit D of 
one of the angles of this face, and 
through the middle E of the oppo- 
site side EC, a line de ; and take 
upon this line a point F two-thirds 
of the distance from the summit 
of the angle or one-third from the 
base ; then draw the line AF. This being done, if we 
conceive the pyramid to be divided into an infinite num- 
ber of sections by planes parallel to the face dcb, all 




104 STATICS. 

these sections will be similar to this face, and they will 
he met by the line af in points, which, being situated 
upon each of them in the same manner as the point F 
is in the face BCD, will be the partial centres of gravity 
of these sections ; hence the centre of gravity of their 
system, which will be that of the solidity of the pyra- 
mid, will be upon the line AF (30). 

For the same reason, having determined the centre 
of gravity a of the area of another face abc, which is 
done by drawing the line ae, and taking upon this line 
the part Ea=f ae ; if through this point, and through 
the summit D of the opposite angle of the pyramid, the 
line Da be drawn, this line will also contain the centre 
of gravity of the solidity of the pyramid. 

Hence the lines af, do, both containing the centre 
of gravity of the pyramid, will necessarily intersect in 
a certain point H ; and the point of intersection of these 
two lines will be the centre of gravity required. 

Remarh I. 

110. It might be demonstrated, independently of the 
consideration of the centre of gravity of the pyramid, 
that the lines af, dg necessarily intersect in one point ; 
for these lines are in the same plane, which is that of 
the triangle ade. 

Remarh II. 

111. Any one of the six edges of a triangular pyra- 
mid being cut by four others ; the fifth which does not 
meet it is called its opposite : if we join the middle 
point of one of the six edges with that of its opposite 



CENTRE OF GRAVITY. 



105 



by a line, it may be demonstrated that the middle of 
this line is the centre of gravity of the pyramid. (See 
the Correspondence of the Polytechnic School^ tome 11, 
page 1.) 

Corollary I. 

112. If from the summit A of one of 
the angles of a triangular pyramid^ 
and through the centre of gravity F 
of the area of the opposite face BCD, 
a line AF he draivn, the centre of 
gravity H of the solidity of the fyra- 
mid will he upon this line^ and at 
onefourth of the distance from the 
face^ or at three-fourths of the dis- 
tance from the summit of the angle. 
Draw the line gf, which will be parallel to ad, be- 
cause the lines ea, ed, are cut proportionally in G, F ; 
the triangles ahd, fhg, whose corresponding angles are 
equal, will be similar, and will give 

AH : HF : : AD : gf. 

But the similar triangles aed, gef, give 

AD : GF : : ED : EF, or : : 3 : 1 (103) ; 

hence, we shall have 

AH : HF : : 3 : 1 ; 




that is to say, ah=3hFj and consequently hf=Jaf, and 



106 



STATICS. 



Corollary II. 

113. It might be demonstrated, in a manner analo- 
gous to that of No. 104j that the distance of the centre 
of gravity of the solidity of a triangular pyramid from 
any plane, is equal to the fourth of the sum of the dis- 
tances of the summits of the four angles of the pyra- 
mid from the same plane. 



Corollary III. 

114. The centre of gravity of the 
solidity of a pyramid abcdef with 
any base is upon the line AG, drawn 
from the summit A to the centre of 
gravity a of the area of the base, and 
at a distance of one-fourth of this 
line from the base or threefourths 
from the summit. 

Conceive the pyramid to be divi- 
ded into an infinite number of sec- 
tions by planes parallel to the base : 
all these sections will be similar to the base, and the 
point where each of them is intersected by the line Aa 
will be situated upon this section in the same manner as 
the point G is upon the base ; consequently this point 
will be the centre of gravity of the section : hence the 
centres of gravity of all the sections will be upon the 
line AG ; hence the centre of gravity of their system, 




CENTRE OF GRAVITY. 107 

which is that of the solidity of the pyramid, will be 
also upon this line (30). 

Moreover, let the base be divided itito triangles by 
the diagonals be, bd, and conceive that through these 
diagonals and through the summit A, the planes abe, 
ABD be drawn, which will divide the proposed pyra- 
mid into as many triangular pyramids as there are tri- 
angles in the base ; then through the centres of gravity 
H, I, K of the triangular bases, draw the lines ah, ai, 
AK ; finally, let the points L, M, N be taken upon these 
lines, upon each of them at the distance of one-fourth 
of its length from the base ; these points will be the 
centres of gravity of the triangular pyramids (112). 
This being done, the points L, M, N, which will divide 
proportionally the lines ah, ai, ak, drawn from the sum- 
mit of the pyramid upon the base, will be in the same 
plane parallel to the base ; hence the centre of gravity 
of the system of triangular pyramids, — that is to say, 
the centre of gravity of the solidity of the proposed 
pyramid, — will be in this same plane ; hence the centre 
of gravity, being found both in this plane and in the 
line Aa, will be at the point of their intersection o. 

Now, the line AG will be cut by the plane lmn in parts 
proportional to the divisions of the lines ah, ai, ak ; 
hence the centre of gravity of the solidity of the 
pyramid will be placed upon AG, at one-fourth of this 
line from the base, or three-fourths from the summit. 



108 



STATICS. 

Corollary IV. 



115. The centre of gravity of the solidity of a cone 
of any base is upon the line drawn from the summit to 
the centre of gravity of the base, and at the distance 
of one-fourth of this line from the base, or at three- 
fourths from the summit ; for this solid may be consid- 
ered a pyramid whose base has an infinite number of 
sides. 



PROBLEM. 

116. To find the centre of gravity of the area of a 
section made in the hull of a vessel hy a horizontal 
'plane. 

Fig- 46, Solution. Let omihec be 

the proposed section, ab the 
^ line of intersection of the 
'% plane of this section with the 
vertical plane drawn through 
the keel of the vessel. It is 
evident, since the whole section is symmetrical on each 
side of the line ab, the required centre of gravity K 
will be upon this line ; thus, to construct this point, it 
will suffice to know its distance ak from a line Cc^ drawn 
through a given point perpendicular to ab. 

For this purpose, let the line ab be divided by the 
perpendiculars or ordinates D(?, E^, f/, . . . into a suffi- 
ciently great number of equal parts, so that the arcs CD, 
DE, EF, .... included between tNvo adjoining perpendi- 



^""""^^ 


\ ^^Si 






\ 1 ^ 








--^ 



CENTRE OF GRAVITY. 109 

culars, may be regarded as riglit lines, which will divide 
the area of the section into trapeziums ; then let each 
of these trapeziums be divided into triangles, by means 
of the diagonals ccZ, De, e/, . . . . This being done, if 
we take the sum of the moments of all the triangles re- 
ferred to the vertical plane passing through the line Cc, 
and divide this sum by the sum of the areas of the tri- 
angles, the quotient will be the distance required AK 
(77). Now, each triangle may be considered as having 
for base one of the perpendiculars, and for height the 
common distance from each other of two consecutive 
perpendiculars ; hence the area of each triangle will be 
equal to the half of the product of the ordinate which 
serves for base, multiplied by the common distance. 
For example, the area of the triangle D'Ee will be equal 
to the half of the product e^XLM ; that of the triangle 
Dde will be the half of dcZxlm, and so of the others. 
Moreover, the distance of the centre of gravity of each 
of the triangles from the plane cc, will be equal to one- 
third of the sum of the distances of the summits of its 
three angles from the same plane (104) : for example, 
the distance of the centre of gravity of the triangle 
DE6 from the plane cc will be one-third of al+am+am, 
and so of the others. 

Hence it will be easy to have the sum of the areas of 
all the triangles, and the sum of the moments of these 
areas referred to the plane C(?; and by dividing the 
second of these two sums by the first, we shall have the 
required distance of the centre of gravity K from the 
line Cc, 

The preceding solution is not rigorous, because the 
parts CD, DE, . . ., cdj c?^, . . ., of the sides of the sec- 

10 



110 STATICS. 

tion are not right lines, as we have supposed ; but it is 
evident that the result will approach exactness as much 
more as these parts are smaller : that is to say, as the 
number of perpendiculars are greater. 

117. The operation just described is susceptible of 
some reduction. Thus, according to the preceding, the 
area of the triangle 



c^ 

C(?cZ=ALX— 5 



That of CDcZ=ALX— , 

A 



That of DcZ^=ALX-^, 

A 



That of DEg=ALX-7r-, 

2 



That of E^/=ALX— , 

A 



That of Er/=ALX-~ ; 

A 



and so on with the others. By adding all these pro- 
ducts together, we see that their sum is equal to the 
product of the common factor al, multiplied by half the 
sum of the two extreme perpendiculars and the sum of 
all the others. 



CENTRE OF GRAVITY. Ill 

As to the moments of these triangles referred to the 
plane cc^ we have 

mi . P 7 C(? Ial 

That of Ced=ALX — X-^, 



That of CD(i=ALX— X-Q-, 



That of D6Zg=ALX^r-X-^7-, 

2 3 



That of DE6 = ALX— x^^— 5 



That of e^/=alx— x-^, 

A o 



That of ef/=alx-^x-^, 
A o 



and so on ; in which we see that the number which mul- 
tiplies AL, in the moment of the last triangle, is always 
equal to three times the number of the intervals minus 
unity ; or, which is the same thing, to three times the 
number of the last perpendicular, less 4. By adding 
together all these moments, we find their sum equal to 
the product of the common factor alxal, multiplied by 
the sum composed of one-sixth of the first perpendicu- 
lar, one sixth of the last, multiplied by three times the 
number of perpendiculars less 4, then of the second 
perpendicular, double the third, three times the fourth, 
. . . and so on. 



112 STATICS. 

Now the sum of the moments and that of the areas 
having the common factor al^ their quotient will also be 
the same if we suppose this factor to be in both terms 
of the division ; hence, to obtain the distance of the 
centre of gravity K from one of the extreme ordinates 
cc, it is necessary^ Ist, to take one-sixth of the first or- 
dinate Go ; one-sixth of the last hA, multiijlied hy three 
times the number of ordinates^ less 4 ; then the second 
ordinate^ double the third^ three times the fourth, . /. 
and so on ; which will give the first sum : 2di, to the 
half of the two extreme ordinates add all the interme- 
diate ordinates ; this will give the second sum : Sd, di- 
vide the first of these two sums by the second^ and multi- 
ply the quotient by the commsn interval of the ordinates. 



PROBLEM. 

118. To find the centre of gravity of the volume of 
the submerged part of the hull of a vessel. 

Solution. We will suppose that the vessel, being 
afloat, has its keel horizontal, and that the vertical plane 
drawn through the keel divides the volume of the hull 
into two perfectly symmetrical parts. This being done, 
the centre of gravity of the part submerged will be in 
this plane, and the question will be reduced to find the 
distances of this point from two lines of known position 
in the vertical plane. 



CENTRE OF GRAVITY. 
Fig, 47. 



113 



B 









r^ 


B 


. 


j>* 


~\ 




5 


/ 


aL.. 


- ^-^-^'-''''^r^ 




/ 


I 




\ 




^ 





f 



■^r 



j..^ 



Let ABODE be the section of a vessel through the ver- 
tical plane CD, its keel, and conceive the plane of flota- 
tion, or the section made in the vessel at the level of the 
water, to be represented by the line b5 parallel to the 
keel. Let the interval of the two lines b6, cd, be divi- 
ded into a certain number of equal parts bb', b'b'', b'^b'^' 
. . . ., and through each point of division suppose there 

are horizontal sections represented by b'J^, '^"h" 

In like manner let the line b5, from the point B of the 

stern-post, be divided into equal parts be, fe', eV ; 

and through each point of division imagine vertical 
planes to be arranged perpendicular to the keel, and 
represented by the lines bb% e/, Vf , . . . ; the sub- 
merged part of the hull will be divided into rectangular 
prisms, whose sides will be perpendicular to the vertical 
plane drawn through the keel, and which will be termi- 
nated on both sides at the surface of the vessel. (It is 
necessary that the divisions of the lines bb'^', b5 should 
be so small that the part of the surface of the vessel, 
which terminates each prism, may be regarded as a 
plane). Finally, let each rectangular prism, represented 
by its base lmnp, be divided into two triangular prisms, 
by a diagonal plane, represented by MP. 

This being done, 1st, each triangular prism will always 
be divided into three pyramids of the same base as the 

10* 



114 STATICS. 

prism (Legendre's Geometry, Book VI.)j and each of 
■which will have for height one of the sides of the prism : 
hence, if, by actual measurement of the vessel, we got 
the length of all the sides, it will be easy to find the 
solidity of each pyramid, by multiplying the area of the 
common base, lmp, by one-third of the side, ^vhich 
measures the height of the pyramid ; and by taking 
the sum of all these solidities^ w^e wall obtain that of the 
submerged part of the hull. 2d. The moment of a tri- 
angular pyramid referred to a plane, being equal to the 
product of the solidity of the pyramid, multiplied by 
one-fourth of the sum of the distances of the summits 
of its four angles from this plane (113), it will be easy 
to find the moment of each pyramid referred to the ver- 
tical plane bb^^^ or to the horizontal plane CD ; because 
the distances of the summits of these angles from each 
of these planes are know^n ; and by taking the sum of 
all these moments, w^e shall have the moment of the 
submerged part of the hull. 

This being done, the quotient of the sum of the mo- 
ments referred to the vertical plane bb'^', divided by the 
sum of the solidities, will be the distance KX of the re- 
quired centre of gravity from the vertical bb^^' ; in like 
manner the quotient of the sum of the moments re- 
ferred to the horizontal plane CD, divided by the sum of 
the solidities, Avill be the distance ky of the same point 
from the keel. We shall have, therefore, the distances 
of the centre of gravity from two lines of known posi- 
tion in the vertical plane drawn through the keel, and 
consequently the position of this point will be deter- 
mined. 



CENTRE OF GRAVITY. 115 

The preceding solution is not rigorous ; because the 
surface of the vessel being curved, the part of this sur- 
face which terminates each triangular prism cannot be 
regarded as a plane, as we have supposed ; but the re- 
sult will approach exactness as much more as the num- 
ber of divisions, both in the direction of the height of 
the vessel, and in that of its length, are greater. 

119. The operation just described is susceptible of 
some reduction; and by reasoning as in No. 116 we find, 
that to get the distance KX of the centre of gravity of 
the submerged part of the hull from the vertical plane 
bb'^^, it is necessary^ 1st ^ for each horizontal section^ to take 
one-sixth of the first ordinate tvhich is in the plane BB^^', 
one-sixth of the last, multiplied hy three times the num- 
her of ordinates contained in the section, less 4 ; then the 
second ordinate, double the third, three times the fourth, 
u . ., which will form a partial sum for each section; 
then add together the half of the first of these sums, the 
half of the last, and all the intermediate ones, ivhich 
will form a dividend ; 2d, to one-fourth of the four or- 
dinates placed at the angles of the rectangle ^W"^'", add 
one-half of all which are upon the contour of this rect- 
angle, and the whole of all those in the interior, which 
will form a divisor ; 3c?, divide the dividend hy the di- 
visor, and multiply the cj^uotient hy the interval bf pa- 
rallel to the distance required KX. 

To find the distance ky of the centre of gravity from 
the horizontal plane drawn through the keel, it is ne- 
, cessary to operate upon the vertical sections as upon 
the horizontal sections in the preceding case : that is to 
say, 1st, for each vertical section, take one-sixth of the 
lower ordinates, c^ie-sixth of that which is in the p)lane 



116 STATICS. 

of flotation^ multiplied hy three times the number of 
ordinates of the section, less 4 ; then the second ordinate 
from the bottom, double the third, three times the fourth^ 
. . . which will form for each section a partial sum ; 
then add together the half of the first of these sums, the 
half of the last, and all the intermediate ones, which will 
form a dividend ; 2(1, divide this dividend by the same 
divisor as in the preceding case, and multiply the 
quotient by the interval BB parallel to the distance 
sought KY. 



RemarTc. 

120. In the preceding problem, the only object is to 
find the centre of gravity of the volume of the submerged 
part of the hull, or, which is the same thing, the volume 
of water displaced by the vessel. But if it were re- 
quired to find the centre of gravity of the vessel itself 
either laden or unladen : that is to say, to find the dis- 
tances of this point from the horizontal plane drawn 
through the keel, and from the vertical plane perpen- 
dicular to the keel, it would be necessary to take, with 
reference to each of these planes, the sum of the mo- 
ments of all the parts which compose the vessel and its 
load, and then to divide each of these sums by the total 
weight of the vessel and its load ; observing, in taking 
the moments, to multiply, not the volume, but the weight 
of each part, by the distance of the partial centre of 
gravity of this part from the plane to which the moments 
are referred ; and the quotients of these divisions would 
be the distances required. 



CENTRE OF GRAVITY. 117 

It will be easy to find, at least in a manner sufficiently 
near, the partial centre of gravity of each of the parts 
of the vessel and its load ; because we may always de- 
compose this part into parallelopipedons, cylinders, py- 
ramids, or other solids of which we have given the 
means of finding the centres of gravity. 



118 



CHAPTER FOURTH. 



ON THE EQUILIBRIUM OF MACHINES. 

121. Every instrument intended to transmit the ac- 
tion of a determined force, to a point which is not found 
upon its direction, so that this force may move a body 
to which it is not immediately applied, and move it in 
a direction different from its own, is called a machine. 

122. In general, we are not able to change the direc- 
tion of a force, but by decomposing this force into two 
others ; one of which is directed toward a fixed point, 
which destroys the force by its resistance, and the other 
acts in a new direction : this latter force, which is the 
only one that can produce any effect, is always a com- 
ponent of the first ; and may be either smaller or great- 
er than it is, according to circumstances. By changing, 
in this manner, the directions and intensities of the 
forces, we may, by the aid of a machine and the points 
of support which it presents, produce an equilibrium 
between two unequal forces which are not directly 
opposed. 

123. The force, whose direction is to be changed by 
employing a machine, is ordinarily named a power^ and 
the term resistance is applied to the body it has to 



MACHINES. 119 

move, or to the force with which it has to produce an 
equilibrium by means of the machine. 

124. We propose here to find only the ratios which 
should subsist between the power and the resistance ap- 
plied to the same machine, in order that, with regard to 
their directions, they may be in equilibrium. We w^ill 
leave friction out of consideration : that is to say, the 
difficulties Avhich the different parts of the machine may 
experience in slipping or rolling upon each other ; 
and we will suppose that cords, when they enter into 
the composition of the machine, are perfectly flexible. 
Thus, having given to a power the intensity which is 
proper for the condition of eqailibrium in this supposi- 
tion, it would not suffice to augment it by a small quan- 
tity to destroy the equilibrium and put the machine in 
motion ; but we must first apply the whole quantity re- 
quired to overcome the obstacles just mentioned ; and 
then a slight increase will produce motion. 

125. Although the number of machines is very great, 
we may regard them all as composed of three simple 
machines, which are : cords ^ the lever^ and the inclined 
plane. We will content ourselves with presenting the 
theories of these three machines, and of those which 
are immediately derived from them; it will then be 
easy, by simple applications, to find the ratio of the 
power to the resistance, for the condition of equilibrium, 
in every machine, however complicated it may be. 



120 



STATICS. 



Article I. 



On the equilibrium of forces ivTiich act upon each other 
by means of cords. 

126. We will suppose that the cords are without 
weight ; and because the property they have of trans- 
mitting force, is independent of their size, we will sup- 
pose them to be reduced to their axes, and regard them 
■^^^- ^^' as straight, flexible, and in- 

extensible lines. Taking 
this for granted, let us con- 
sider, first, the case of equi- 
librium between three forces 
p, Q, R, acting upon each 
other by means of three 
cords united together by a 
knot A. 

1st. The three forces p, 
Q, R, cannot be in equili- 
brium, unless the three di- 
rections, and consequently 
the cords by means of which they transmit their actions, 
are in the same plane (10). 

2d. If we represent any two of these forces, for ex- 
ample, the forces P, Q, by the parts AC, ad of their di- 
rections, and upon these lines as adjacent sides construct 
the parallelogram ACBD, the diagonal ab will represent 
in intensity and direction the resultant of these two 
forces (36) ; now the three forces being in equilibrium, 
the force R must be equal and directly opposed to the 
resultant of the two others ; hence the direction of the 




CORDS. 121 

force R "will be in the prolongation of BA, and its in- 
tensity will be represented by this diagonal : thus we 
shall have 

p : Q : R : : AC : AD : AB ; 

or because the sides ad, bc of the parallelogram are 
equal to each other, the three forces P, Q, R will be to 
each other as the sides of the triangle ABC. 

127. The angles of the triangle abc, being given by 
the directions of the forces P, Q, R, and the magnitudes 
of its sides being proportional to the intensities of these 
forces, it follows that, of the six things to be considered 
in the equilibrium in question, namely, the directions 
of the forces and their intensities, any three being 
given, we can find the three others in all cases in which, 
of the six things to be considered in the triangle ABC, 
namely, the angles and the sides, three being given, we 
can determine the three others. 

For example, when the three forces P, Q, R are known, 
we can find the angles which the cords make with each 
other when in equilibrium, by constructing the triangle 
ABC, the sides of which are proportional to these forces. 
But when the directions are given, we can know only 
the ratios of the three forces ; because in the triangle 
ABC, the knowledge of the three angles determines only 
the ratios of the sides, but does not determine their 
magnitudes. Thus, it will be necessary, besides, to know 
the magnitude of one of the three sides p, Q, R, in order 
to find that of the two others, by means of the propor- 
tional series : 

P : Q : R : : AC : BC : AB. 
11 



122 



STATICS. 



Remark I. 

^'9' 49. 128. If the three cords be 

united by a slip-knot ; for ex- 
ample, if the cord paq passes 
through a ring attached to 
the extremity of the cord ra, 
the conditions just enuncia- 
ted are not sufficient to es- 
tablish equilibrium : it is ne- 
cessary, in addition, that the 
angles pab, qab, formed by 
the two parts of the cord and 
by the prolongation ab of the direction of the other 
cord, should be equal ; for it is evident that, otherwise, 
the ring A would slip upon this cord toward the greater 
of the two angles. 




Remark II. 

129. What has just been said contains the whole 
theory of equilibrium between three powers applied to 
cords united in the same knot ; but we have supposed 
the construction of the parallelogram acbd, we may 
however enunciate this theory independently of all con- 
struction. 

Thus, in every triangle abc, the sides are proportional 
to the sines of the opposite angles ; that is to say, w^e 
have 



AC : BC : ab : : sm abc : sin bag : sin acb. 



CORDS. 



123 



N0W5 t^^ sines of these angles are respectively the 
same as those of their supplements raq, rap, paq; 
hence we have also 

AC : EC : AB : : sin raq : sin rap : sin paq, 

and consequently 

p : Q : R : : sin raq : sin rap : sin paq. 

That is to say, when the three powers^ which act hy 
cords upon the same knot^ are in equilibrium^ each of 
them is as the sine of the angle formed hy the directions 
of the other two. 

Corollary L 




Fig^ 50. 130. If the cords ap, aq, 

instead of being drawn by 
two powers, be attached to 
two fixed points at P and Q, 
and we represent the force 
R by the diagonal AB of the 
parallelogram abcd, the two 
sides AC, AD will represent 
the tensions of these two cords, or the efforts which 
they exert upon the fixed points in the line of their di- 
rections. 



Corollary II. 

131. When the angle paq is very large, the sides AC, 
AD of the parallelogram are very great compared with 
the diagonal AB, and consequently the effort which the 



124 



STATICS. 



power K exerts upon the fixed points p, q, to make them 
approach each other, is very great compared with this 
power. Hence, by means of cords, we can put a 
moderate power in the condition of producing a very 
great effect. 

Corollary III. 

132. In the case of equilibrium, however small the 
force R may be, the diagonal AB, which represents it, is 
not zero, and the three points c. A, d, are not in a straight 
line ; hence, by supposing that a cord paq without weight 
is stretched in a line by two forces p, q, the smallest 
force R, applied to A, will bend it at this point and cause 
it to make an angle paq. Thus, it is rigorously impossi- 
ble to stretch a heavy cord in a straight line, unless it 
be vertical ; for the weight of the parts which compose 
it may be regarded as forces applied to this cord, and 
which must necessarily deflect it from a straight line. 



Corollary TV. 



Fig, 51. 




133. If any number 
of powers p, q, r, S, 
T, . . . act upon each 
other by cords join- 
ed together three by 
three in the same knot. 



it is easy J 



from what 



precedes, to find the 
ratios which these pow- 



CORDS. 125 

ers should have with each other as to their directions, 
so as to be in equilibrium. For the general equilibrium 
cannot take place, unless, 1st, the three powers joined 
together in each knot are in equilibrium with each other ; 
2d, each of the cords ab, bc, which join two knots, are 
equally stretched in the two directions. Hence, by 
naming y, x, the tensions of the two cords ab, bc, we 
will have (129), by reason of the equilibrium at the 
knot A, 

P : Q : : sin qab : sin pab, 
p : V : : sin qab : sin paq ; 

by reason of the equilibrium at the knot b, 

V : R : : sin Rbc : sin abc, 

V : X : : sin rbc : sin abr ; 

by reason of the equilibrium at the knot c, 

X : S : : sin SCT : sin bct, 
X : T : : sin SCT : sin BCS. 

And by continuing these proportions, we will find the 
ratio of any two of these powers, and the ratio of one 
of them to the tension v, x of any cord whatever. 

For example, by multiplying in order the 2d propor- 
tion and the 3d, we find 

p : R : : sin qab x sin rbc : sin paq x sin abc; 

by multiplying the 2d and 4th, 

11* 



126 STATICS. 

p : X : : sin QABXsin rbc : sin PAQXsin abr ; 

by multiplying the 2d, 4th and 5th, 

p : S : : sin QABXsin rbc X sin SCT 
: sin PAQXsin abr x sin bct; 

by multiplying the 2d, 4th and 6th, 

p : T : : sin QABXsin rbc x sin scT 
: sin PAQXsin abr x sin bcs, 

and so on. 

It also follows from this, that the three cords united 
by the same knot are in the same plane (126), although 
those which are united at two knots may be in diflferent 
planes. 



Corollary V. 




134. If the forces Q, R, S, 
be weights suspended by the 
knots A, B, c, to the same 
cord EABCF, and this cord 
be retained at its extremities 
by two fixed points E, P : 

1st. The main co^d and the 
cords of the weights Q, R, S, 
are in the same vertical plane ; 



CORDS. 127 

for the two parts ea, ab of the cord are in the vertical 
plane drawn through the cord AQ ; in like manner, the 
two parts AB, BC are in the vertical plane drawn through 
BR : now these two vertical planes pass through the 
same line ab, and coincide ; hence the parts EA, AB, BC 
of the cord, and the directions of the cords AQ, BR, are 
in the same vertical plane. In the same manner it may 
be demonstrated that the part CE of the cord and the 
direction CS are in this same plane, and so on. 

2d. The tensions of the two extreme parts of the cord 
are to each other reciprocally as the sines of the angles 
which these parts make with the vertical ; for the angles 
QAB, ABR, are supplements of each other, and have the 
same sine : it is the same with the angles rbc, bcs, and 
so of the rest ; hence, by neglecting the common factors 
in the proportion which gives the ratio of the two ex- 
treme tensions p, t (183), we have 

p : T : : sin sct : sin paq. 

3d. The vertical hi, drawn through the point of inter- 
section a of the prolongations of the two extreme parts 
of the cord, passes through the centre of gravity of the 
system of all the weights Q, r, s, . . . ; for the two ex- 
• treme parts being in the same plane, their tensions have 
a resultant whose direction passes through the point a ; 
moreover, these tensions supporting the system of 
weights Q, R, S, . . . their resultant should be vertical, 
and pass through the centre of gravity of these weights ; 
hence the point a and the centre of gravity of the 
weights Q, R, s, are in the same vertical. 



128 



STATICS. 



Corollary VI. 




^^9'^^' 135. When a heavy cord 

EHF is suspended in equilibri- 
um to the two fixed points 
Ej F, we may consider its axis 
as a thread without weight, 
charged with a weight dis- 
tributed throughout its whole 
extent : hence Ist, this axis is 
in the vertical plane drawn through the two points of 
suspension; 2d, if the directions of the two extreme 
elements of this axis be prolonged to eg. Fa, and through 
the point of intersection a we draw the vertical IH, the 
tensions of these two elements are to each other recipro- 
cally as the sines of the angles which these elements 
make with the vertical : that is to say, by naming p and 
T these tensions, we have 



p : T : : sm igf : sm IG e ; 

3d, the centre of gravity k of the cord is in the ver- 
tical IH. 

Finally, by considering the total weight z of the cord 
as a force applied to the point a of its direction, we will 
find the efi*orts which the cord makes upon the two 
points of support E, F, along the directions eg, fg, by 
decomposing the force z into two others which act in 
these directions ; and we shall have (129) 



z : P : T 



sin EGF : sm igf : sm ige. 



CORDS. 129 



RemarJc. 



136. So far we have supposed that there were only 
three cords united at each knot, because, if the cords as- 
sembled at the same knot are greater in number and 
included in the same plane, it is not sufficient to know 
their directions in order to find what ratios the applied 
powers should have so as to be in equilibrium : that is 
to say, these ratios may vary in an infinite number of 
ways, without the forces ceasing to be in equilibrium. 

Thus, whatever may be the number of powers directed 
in the same plane, it suffices, in order to be in equili- 
brium about the same knot, that the resultant of any 
two of them is equal and directly opposed to the re- 
sultant of all the others ; hence all these forces, except 
two, being taken at pleasure, which determine the in- 
tensity and direction of the resultant, we can find the 
intensities of the latter two forces which Avill make an 
equilibrium with this resultant. 

However, when four cords, joined at the same knot, 
are not in the same plane, their directions being given, 
the ratios of the intensities which the applied forces 
must have, to produce an equilibrium, are determined : 
for we have seen (44), that these forces must be to each 
other as the diagonal, and the adjacent sides of the 
parallelopipedon constructed on their lines of direction. 
But when the forces are not directed in the same plane, 
and their number is greater than four, the ratios of the 
forces are no longer determined by the knowledge of 
the direction of the cords. 



130 



STATICS. 



Article IL 



On the Equilibrium of the Lever. 

Fig^ 54. 137. The lever is an 

inflexible rod acb {Fig. 
54)j or cab {Fig, 55), 
either straight or curved 
and moveable around 
one of its points c, ren- 
dered fixed by means of 
any obstacle ; and this 
obstacle is termed the 
'point of support or ful- 
crum, 

138. First, supposing 
the lever to be without 
weight, and that it cannot 
in any manner slip upon 
its fulcrum, let p, q be two 
powers applied, cither im- 
mediately or by means of 
the cords ap, bq, to the 
two pomts A, B of a lever. 
If we consider the resist- 
ance of the point c as the effect of a third force r ap- 
plied to the lever at this point, we have seen, in order 
that equilibrium may subsist between these three forces, 
1st, their directions must be included in the same plane, 
and meet in the point D (10) ; 2d, the forces p, q must 
be to each other reciprocally as the perpendiculars CE, 




THE LEVER. 131 

CF, let fall from the fulcrum upon their directions (35) : 
that is to say, we must have 

p : Q :: CF : ce; 

3d, if we lay aff from the point D the lines dl, dm, 
upon the directions of the forces p, Q, proportional to 
the intensities of these forces, and finish the parallelo- 
gram DLMN ; the diagonal dn will represent in intensity 
and direction the force R, and consequently the resist- 
ance of the fulcrum (35) ; thus we will have 

p : Q : R : : DL : DM or NL : DN, 
or (129), 

p : Q : R : : sin qdr : sin pdr : sin pdq. 

Corollary I. 

139. If we leave out of consideration the resistance 
of the fulcrum : that is to say, if we suppose this point 
to be capable of an indefinite resistance, it is necessary, 
in order that the two powers p, q may be in equilibrium 
around this point by means of the lever, 1st, their di- 
rections and the fulcrum should be in the same plane ; 
2d, the two forces p, q should tend to turn the lever 
around the fulcrum c in opposite directions, and their 
moments, referred to this point, should be equal : that 
is to say, we should have (80) 

PXCE=QXCF. 



132 STATICS. 



Corollary IL 

140. Hence it appears, 1st, that however small the 
power Q may be, we may always, by means of a lever, 
put it in equilibrium around a fulcrum c, with another 
power P of given intensity and direction ; for the di- 
rection of the force p being known, the distance CB of 
this direction from the fulcrum will be known, and we 
will know the moment pxce : hence it will be sufficient 
to arrange it so that the moment qxcf of the power, is 
equal to the preceding : that is to say, to direct this 
power in such a manner that its distance CF from the 

P X CE 

fulcrum is equal to , and that it tends to turn the 

lever in the opposite direction to the force P. 

2d. If the distance CF of the direction of the force 
Q from the fulcrum is known, we will find the intensity 
which this force must have in order to produce an equi- 
librium with the force p, by dividing the moment of this 
latter force by the distance CF : that is to say, we shall 
have 



PXCE 

Q= . 

CF 



Corollary III. 

141. The effort or load, which the fulcrum C sustains, 
being equal to the resultant of the two forces P, Q, we 
will find this load by means of the following proportion : 



THE LEVER. 133 

p : Q : R : : sin qdr : sin pdr : sin PDQ : 

that is to say, we shall have 

pxsin PDQ 
sm QDR 

or, 

QXsin PDQ 



R=- 



sm PBR 



Corollary IV. 

142. Hence, if the fulcrum does not possess indefinite 
resistance, in order that it may not be moved, and that 
equilibrium may subsist, it is necessary that the resist- 
ance in the direction CD, should be equal to the result- 
ant of the two forces P, Q : that is to say, equal to 

PXsin PDQ 



or, which is the same, to 



sm QDR 

QXsin PDQ 
sin PDR 



Corollary V. • 

143. In general, of the six things to be considered 
in the equilibrium of the lever, namely the intensities 
and directions of the two powers P, Q, and those of the 
load on the fulcrum, any three being given, the other 
three may be determined in all cases; or of the six 

12 



134 STATICS. 

analogous things which may be considered in the tri- 
angle DLN, namely, the sides and the angles, three being 
given, we can determine the others. 

Remarh I. 

144. If the lever can slip upon the fulcrum, the con- 
ditions, which have just been given, are not sufficient to 
keep the lever stable, and to produce an equilibrium ; 
it is necessary, besides, that the direction DC of the load 
on the fulcrum should be perpendicular to the surface 
of the lever at the point c ; for, if this direction were 
oblique, the lever would have a tendency to slip towards 
the side of the greater angle, and in fact would slip, 
whenever this tendency should be greater than the fric- 
tion upon the fulcrum which opposes this effect, as we 
will demonstrate in treating of the inclined plane. 

Remark II. 

145. What has just been stated contains the whole 
theory of the equilibrium of two powers applied to a 
lever, considered to be without weight, and retained by 
a fulcrum ; we will now make application of it to a few 
simple cases. 

^^9' S6. If the directions of the pow- 

^^^ ers P, Q, applied to a lever, are 

b^.-..^^^^^^^--|e' parallel to each other: for ex- 

I ample, if there are two weights 

I suspended at the points A, B, 

® the load, which the fulcrum c 
sustains, is equal to their sum 



THE LEVER. 135 

P+Q.5 and the two perpendiculars ce, cf, let fall from 
the fulcrum upon their directions, are in a straight line. 
Hence, if the lever be straight, the triangles ace, bcp 
will be similar, and give 

CF : ce : : CB : CA. 

Hence we shall have, in the case of equilibrium, 

p : Q : : CB : CA : 

that is to say, the weights P, Q will be to each other re- 
ciprocally as the arms of the lever. 

Thus, the intensity and the lever arm of a resistance 
p being given, 1st, the lever arm, which must be given 
to a power Q in order to produce an equilibrium with it, 
will be 



PXCA 

CB= 



2d. The intensity of the power necessary to be ap- 
plied to the given point B, in order to produce an equi- 
librium with it, will be 

PXCA 

Q= . 

CB 

Finally, if the two weights P, Q, and the length AB 
of the lever, be given, we will find the fulcrum c, around 
which these weights will be in equilibrium, by dividing 
the lever ab into two parts reciprocally proportional to 
the two weights. 



136 STATICS. 



Bemarh III. 

146. When there are more than two powers applied 
to the same lever, it is not sufficient to know their di- 
rections, and the position of the fulcrum, in order to 
determine the ratios which they should have to each 
other so as to be in equilibrium ; but, as equilibrium 
cannot take place between several forces about a ful- 
crum, unless the resultants of all the forces are destroyed 
by the resistance of this point, it is evident, in this 
case, that the conditions of equilibrium are reduced to 
the two follovfing : 1st, all the forces must have a single 
resultant ; 2d, the direction of this resultant must pass 
through the fulcrum. 

Corollary. 

147. If the directions of all the forces are included 
in the same plane, these forces have necessarily a single 
resultant (43), and the first condition is fulfilled ; hence 
it suffices for the equilibrium, that the direction of this 
resultant passes through the fulcrum, or, which amounts 
to the same, the sum of the moments of the forces, 
which tend to turn the lever in one direction around the 
fulcrum, is equal to the sum of the moments of those 
which tend to turn it in the opposite direction. 

Remark IV. 

148. So far we have abstracted weight from the 
lever; but if this weight enter into consideration, it 
must be regarded as a new force, applied to the centre 



THE LEVER. 



137 



Fig. 57. 



of gravity of the lever in a vertical direction ; and in 
the case of equilibrium, the conditions just stated sub- 
sist between all the forces, including that under con- 
sideration. 

Let p, Q be two weights sus- 
pended from a heavy lever ab, 
and in equilibrium about the 
fulcrum c : we will consider the 
weight of the lever as a third 
weight S, suspended from the 
centre of gravity of the lever, 
and the sum of the moments of the two weights Q, S, 
referred to the fulcrum c, will be equal to the moment 
of the weight p : that is to say, we shall have 




QXCB-f-sxca=PXCA; 

or, subtracting from these two equal quantities the mo- 
ment of the lever sxca, 

QXCB=pxcA— sxca. 

Thus, knowing the length and the weight of the lever, 
the position of its centre of gravity, that of the fulcrum 
and one of the two weights P, Q, it will always be pos- 
sible to get the other weight ; for we shall have 



p— 



QXCB-fSXCa 



CA 



and 



Q=- 



pxcA— sxca 

CB 
12* 



138 



STATICS. 



The load at the fulcrum is evidently equal to the sum 
of the weights p-f q+s. 

Fi9' 58. 149. But, if the weight p, 

suspended from the heavy 
]g lever ab, be retained in equi- 
librium about the fulcrum c, by 
means of a vertical power Q, 
and directed upwards, the mo- 
ment of the force Q, which 
tends to turn the lever in one 
direction, will be equal to the 
sum of the moments of the 
weights P, S, which tend to 
turn it in the opposite direc- 
tion, and we shall have 

QXCB=PXCA + SXCa, 

or, subtracting the moment of the lever, 

QXCB— ■SXCG=PXCA. 

Thus the intensities, which the two forces P, Q must 
have in order to be in equilibrium, will be 




P= 



QXCB— sxca 



and 



CA 



pxcA+sxca 

CB ^ 



and the weight upon the fulcrum p+s— Q. 



THE LEVER. 



139 



C'' ROLLARY. 




Ftg. 57. 150. Hence it appears by 

regarding the weight P as a re- 
sistance, and the force Q as a 
power which has to bring the 
weight into equilibrium or to 
overcome it, by means of the 
lever AB, the weight of this 
lever is a force which may either increase or diminish 
the power, according as this weight tends to turn the 
lever in the same direction as the power, or in the oppo- 
site direction. For example, in the case of Fig. 57, 
the weight of the lever increases the power Q ; and by 
prolonging the lever arm cb of this power, it would be 
placed in the condition of producing an equilibrium with 
a greater resistance, for two reasons : 1st, because its 
^^9' 58. moments would be thus aug- 

mented; 2d, because the 
weight S of the lever would 
be augmented, which alone 
would produce an equilibrium 
with a greater part of the 
resistance. But, in the case 
of I^ig, 58, the weight of 
the lever diminishes the pow- 
er Q, and we cannot increase 
the length of the lever arm 
^ CB without at the same time 

augmenting its weight s, which forms part of the resist- 
ance ; thus, in order that there may be in this case an 






y 



140 STATICS. 

advantage in prolonging the arm of the lever, it is 
necessary that the moment of this prolongation should 
be less than the resulting increase in the moment of the 
power. 



THEOREM. 

■^^^ ^^' 161. Two powers p, 

//\ Q, applied to a lever 

AB5 and in equilibrium 

>f about a fulcrum c, are 

to each other reeipro- 

y- ^^^"^ ^ cally as the spaces 

which these powers tra- 
verse in the line of 
their directions^ if the 
equilibrium be disturbed infinitesimally. 

Demonstration. From the fulcrum c let fall the 
perpendiculars CE, CF upon the directions of the powers ; 
and instead of the straight lever ab, let us consider the 
bent lever ecf ; at the extremities E, r of which con- 
ceive the powers p, Q to be applied ; then suppose, by 
virtue of a derangement in the equilibrium, that the 
bent lever ecf takes the infinitely near position ecf 
This being done, the small arcs E^, f/, will be the spaces 
which the powers p, q would traverse by virtue of this 
derangement. Now, the angle ecf of the bent lever 
being invariable, the two angles ec<?, fc/ are equal, and 
we have 



THE LEVER. 

CF : CE : : f/: e^; 
moreoverj because of the equilibrium, we have (35) 



141 



p : Q : : CF : CE ; 



hence we have also 

p : Q : : f/ : E^. 

We will have occasion to show, in a subsequent part, 
that an analogous proposition occurs in cases of equi- 
librium for all other machines. 



On Pulleys. 



Fig. 60. 




152. A pulley is a 
wheel having a groove on 
its circumference to re- 
ceive a cord pgdhq, and 
is traversed at its centre 
by an axle e, upon which 
it turns in a sheath or 
block EL. 



142 



STATICS. 




II. 

-%• 6^- 153. Suppose the axis 

of the pulley being fixed, 
two forces p, Q are ap- 
plied to the extremities 
of the cordj and this cord 
being perfectly flexible 
exerts no friction upon 
the groove of the pulley, 
so that it may slide free- 
ly on this rim. What- 
ever the figure of the 
pulley may be in other 
respects, that is to say, 
whether the arc gbh, em- 
braced by the cord, be circular or not, it is evident that 
the two forces cannot be reciprocally in equilibrium 
unless they are equal ; for if they were unequal, the 
greater would overcome the smaller by causing it to 
slide in the groove of the pulley. 

Upon the same supposition, the pulley, having no 
other fixed point than its centre, and being drawn by 
the two forces p, q, cannot remain at rest, unless the 
resultant of these two forces is directed towards the 
centre, and is destroyed by the resistance of this point. 
Hence, having prolonged the directions pg, qii of the 
two cords, until they meet in some point A, and taking 
upon these directions the equal lines ab, ac to repre- 
sent the forces p, Q, if the parallelogram abdc be com- 
pleted, the diagonal ad, which will represent the resultant 
of these two forces, must pass through the fixed point E. 



PULLEYS. 



143 




Now, when the figure 
is circular, this last con- 
dition is always fulfilled : 
thus, the triangle abd 
being isosceles, the angle 
BAD is equal to the angle 
BDA, and consequently to 
the angle dac ; hence the 
diagonal AD divides the 
angle bag into two equal 
parts. But, if we draw 
the line ea, this line di- 
vides the same angle into 
two equal parts ; for if, 
through the centre e, we draw, to the points of contact 
of the cords, the radii Ea, eh, perpendicular to the di- 
rections of these cords, the two rectangular triangles 
EaA, EHA will be in all respects equal, and we shall have 
the angle eag of the one, equal to the angle eah of the 
other. Hence the line ea, and the diagonal DA, will 
have the same direction. 

Hence, the centre of a pulley being fixed and its 
figure circular, it is necessary that the two forces P, Q, 
to be in equilibrium, should be equal, and that the pulley 
remain at rest, at the same time, about its axis. 

The load which the axis of the pulley sustains is evi- 
dently equal to the resultant of the two forces P, q; 
hence, if we name R this load, we shall have (36) 



p ; Q : R : : AB : AC : ad. 



144 



STATICS. 



Fig, 60. 




Finally, draw the cord 
GH of the arc embraced 
by the rope ; the two tri- 
angles GHE, ABD will be 
similar, because they will 
have their sides perpen- 
dicular each to each ; and 
we shall have 

AB : AC : AD : : GE : eh : gh; 

hence we shall have 

p : Q : B : : GE : EH : gh. 



III. 

154. If the axis of the pulley is not absolutely fixed, 
but retained simply by the power s, by means of the 
block EL and the cord LS ; in order that this axis may 
be at rest, and the three forces p, q, S in equilibrium, 
it is necessary that the force S should be equal and di- 
rectly opposed to the load which the axis sustains. 
Hence, 1st, the direction of this force should coincide 
with the line ea ; 2d, its intensity should be equal to 
the resultant K of the two forces P, Q, and we shall 
have 



p : Q : S : : GE : EH : GH. 



PULLEYS. 



145 



Thus, when two forces p, q, applied to a rope em- 
hracing a pulley^ are in equilihriiim with each other ^ and 
with a third force S applied to the axis of the pulley^ 
Ist, the two forces P, Q are equal to each other ; 2di^ the 
direction of the force S bisects the angle formed hy the 
directions of the two others ; Sd, each of the two forces 
p and Q is to the third force S, as the radius of the pul- 
ley is to the chord of the arc embraced by the rope. 



Corollary L 



60. 155. We observe that 

if the cord, fastened to 
the block, instead of being 
drawn by a force B, is at- 
tached to a fixed point of 
indefinite resistance, and 
if it be proposed, by em- 
ploying the pulley, only 
to place the two forces 
p, Q in equilibrium, or to 
overcome a resistance p, 
by means of a power Q, 
the pulley does not assist 
the power : it has no 

other efi*ect than to change the direction of this force, 

without altering its intensity. 




18 



146 



STATICS. 



Fig, 61. 




But if one of the ex- 
tremities of the cordj 
which embraces the pul- 
lejj is attached to a fixed 
point P, and, by using 
the pulley, we design to 
place a power Q in equi- 
librium with a resistance 
S, attached to the block, 
the pulley assists the 
power, which is always 
less than the resistance ; 
for we have 



: s : : EH : GH. 



Corollary II. 



Fig, 62. 



Fig. 63. 



156. When the directions 
of tAVO parts pg, qh of the cord 
are parallel to each other, and 
consequently to that of the 
cord of the block, the chord 
GH becomes a diameter, and 
is double the radius ; the pro- 
portional series of No. 153 
then becomes 

p : Q : S : : 1 : 1 : 2, 



that is to say, the force S, or the load of the axis of the 
pulley, is equal to the sum of the two powers p, q, or to 





PULLEYS. 147 

double one of them. Thus, in the case of Fig. 63, the 
power Q, which, by means of the pulley and the point 
of support P, will produce an equilibrium with the re- 
sistance S, w^ill be only half of this resistance. 



IV. 

157. A pulley is said to be immoveable when its 
block is attached to a fixed point and the power and re- 
sistance are applied to the cord which embraces it ; but 
when the resistance is attached to the block, and the 
pulley has to move with it, as in Figs, 61, 63, the pulley 
is said to be 7noveabIe. 

This being established, let there be any number 
of moveable pulleys, {Fig, 64,) and considered without 
weight ; let the first bear a weight P suspended to its 
hook, and embraced by a cord, one of whose extremi- 
ties is attached to the fixed point d, and the other is 
applied to the block of the second pulley ; let this pul- 
ley be embraced by another cord, one of whose extremi- 
ties is fixed to the point H, and the other attached to 
the block of the next pulley ; let this third pulley be 
embraced by a third cord fixed on one side to M, and 
drawn on the other by a power Q ; and so on, if the 
number of pulleys be greater. Finally, supposing the 
whole system to be in equilibrium, draw the radii and 
the subtending chords of the pulleys as is shown in the 
figure. We may consider the equilibrium of the first 
pulley A as though this pulley were alone ; and represent- 
ing by X the tension of the cord BX, we shall have (155) 

p : X : : BC : BA. 




For the same reason, calling Y the tension of the cord 
FY, we shall have 

X : T : : Fa : ef. 

We shall have, in like manner, for the third pulley, 

Y : Q : : KL : IK ; 

and so on, whatever may be the number of pulleys. 
Hence, by multiplying in order all these proportions, 
we shall have 

p : Q : : bcxfgxkl : baxefxik: 



PULLEYS. 



149 



that is to say, the resistance is to the power as the 
product of the subtending chords is to the product of 
the radii. 



Corollary. 



158. When all the ropes CD, gh, lm, . . are parallel, 
the subtending chords will be diameters, and the pre- 
ceding proportion will become 

p : Q :: 2x2x2 : 1x1x1, or :: 8 : 1; 



Fig. 65. 




from which it is evident, that, 
in this case, the resistance is to 
the poAYer as the number 2, 
raised to a power indicated by 
the number of moveable pul- 
leys, is to unity. 

Thus, by suitably increasing 
the number of moveable pulleys, 
we may put a moderate force in 
equilibrium with a very great 
resistance. For example, with 
•three pulleys, and by means of 
the fixed points D, H, M, the 
power produces an equilibrium 
with a resistance eight times 
greater than itself. 

However advantageous this 
disposition of moveable pulleys 
may at first appear, it is rarely 



150 STATICS. 



employed ; because, in order to make the first pulley A 
traverse a certain space, it is necessary that the second 
should traverse a double space, and the third a quadru- 
ple ; and so on, which requires too much room ; and 
thus, most generally, muffles are used instead. 



V, 



159. The term muffle is applied to a system of several 
pulleys assembled in the same block and upon separate 
axes, as in Figs, 66, 67, or upon the same axis, as in 
Fig. 68. A fixed muffle and a moveable muffle are 
always employed at the same time ; and all the pulleys 
of the two muffles are embraced by the same cord, one 
of whose extremities is attached to one of the two 
muffles, and the other extremity is drawn by the power ; 
and the resistance is suspended to the moveable muffle. 

We may give difi^erent diameters to the pulleys, and 
arrange them in such a manner that all the parts of the 
cord, which go from one muffle to the other, may be 
parallel to each other, as in Figs, 66, 67 ; but this ar- 
rangement increases the extent of the muffles. They 
may be reduced to a volume much smaller and more 
convenient, by mounting in each of them all the pulleys 
upon the same axis, as in Fig, 68. By this means, the 
cords which are on one side of the muffles are not 
parallel to those on the other side ; but, when the 
distance of the muffles is considerable, the departure 
from parallelism is very small, and it may be regarded 
as insensible. 



151 



Fig, 66. 



Fig, 68. 




-^-t--^ ^ 





m 




160. By considering, then, tlie cords of the muffles 
as parallel to each other, and abstracting all weight 
from the whole machine, let Q be a power in equilibrium 
with the resistance P, suspended from the block of the 
moveable muffle. Equilibrium cannot exist throughout 
the whole machine, unless it occur in each particular 
pulley, and the two parts of the cord w^hich embraces 
this pulley are equally stretched (154). Thus, the ten- 
sions of the two cords QA, bc are equal to each other ; 
so likewise are those of the two cords bc, be, and those 



152 



STATICS. 



Fig. 66. ^ of the cords DE, FG, and so on, what- 
ever may be the number of the cords ; 
hence all the cords which go from one 
muffle to the other are equally stretch- 
ed. Now the sum of these tensions 
produces an equilibrium with the re- 
sistance P, and is equal to it ; or, what 
is the same, the tension of one of the 
cords, multiplied by their number, is 
equal to the resistance ; hence the ten- 
sion of one of the cords, or the power 
Q, is the quotient of the resistance p di- 
vided by the number of cords which go 
from one muffle to the other. 

Hence it is evident, that, in the case 
of J%. ^^^ where the extremity of the 
cord is attached to the fixed muffle, the 
power Q should be one-sixth of the re- 
sistance to be in equilibrium with it, and 
in the case of Mg, 67, where the ex- 
tremity of the cord is attached to the 
^ moveable muffle, it should be one-fifth, 

because there is one pulley and one cord less. 

If it be desired to introduce the consideration of the 

weight of the moveable muffle, it may be regarded as 

constituting part of the resistance. 



WHEEL AND AXLE. 



163 



Of the Wheel and Axle. 

I. 

161. The wheel and axle^ windlass^ or capstan^ is a 
machine consisting of a cylinder moveable upon an axis, 
and of a cord, which, having one of its extremities 
wound around the cylinder, while a power Q causes it 
to turn, overcomes a resistance P attached to its other 
extremity. The cylinder is furnished at its two ends 
with trunnions A, B, bearing upon supports, which are 
named hoxes^ and by means of which it turns freely on 
its axis. 

Fig. 69. 




162. There are several modes of applying the power 
to this machine, to communicate the movement of rota- 
tion to the cylinder. 



154 STATICS. 

1st. We may unite solidly with the cylinder and upon 
the same axis, a wheel, whose circumference, having a 
groove in it like that of a pulley, is surrounded by a 
second cord. This cord, being drawn by the power, 
causes both the wheel and the cylinder to turn upon 
their common axis. This first arrangement, to which we 
will refer all the others, is rarely employed, because it 
requires the cord of the wheel to be very long, while 
the space to be traversed by the resistance is incon- 
siderable. 

2d. The rim of a wheel, furnished with spokes placed 
equally far apart, and to which men's hands are applied, 
furnishes them with the means of making use of part 
of their weight in turning the machine. 

8d. In other cases, instead of the wheel, there is 
mounted upon the cylinder a large hollow drum in which 
men or animals can walk ; and then by their feet they 
cause both the drum and the cylinder to turn. 

4th. Sometimes, in place of making use of the wheel 
and the drum, the cylinder is traversed by bars perpen- 
dicular to its axis, and at the extremities of which men 
act by the force of their muscles and part of their 
w^eight. 

5th. Finally, most frequently, and when the resist- 
ance is not very great, there are adapted to the ex- 
tremities of the cylinder one or two cranks, which men 
turn by employing the force of their arms. 

163. The names of this machine vary with its object, 
and even Avith its position. Ordinarily it is named wind- 
lass^ and wheel and axle, when the cylinder is horizontal ; 
and capstan, when the cylinder is vertical, and horizon- 
tal bars are used to put it in motion. 



WHEEL AND AXLE. 



155 




It is evident that the different modes in which the 
power may be applied to the cylinder of the wheel and 
axle, can all be referred in theory to the first we have 
described; for, whatever may be the direction of the 
power, when it is directed in a plane perpendicular to 
the axis of the cylinder, we may always conceive it to 
be applied to a wheel whose circumference would be 
tangent to the direction of this power. Thus we will 
suppose that vxyz being the cylinder of the machine, 
whose axis acb is perpendicular to the plane of the 
wheel hDB^Jc, 1st, the power Q is applied to the circum- 
ference of this wheel in any direction dq, contained in 
the plane of the wheel^ and tangent to the circumference 
of the radius CD, at a given point D ; 2d, that the di- 
rection KP of the resistance is tangent at K to the sur- 
face of the cylinder, and situated in a plane parallel to 
that of the wheel. Lastly, to render the conception 
clear, we will suppose that the axis ab of the cylinder 
is horizontal, and consequently that the direction kp of 



156 STATICS. 

the resistance is vertical. This being granted, two 
questions present themselves to be solved : the first is 
to find the ratio Arhich the power Q and the resistance P 
should have to produce an equilibrium ; the second is to 
find the weight which the supports of the two pivots 
Aj B, sustain. 



II. 

164. To solve the first of these two questions, let us 
conceive a horizontal plane kmen to pass through the 
axis AICB of the cylinder : this plane will pass through 
the point K, where the direction of the resistance touches 
the surface of the cylinder ; moreover, it will intersect 
the plane of the wheel in a horizontal line ME, which 
will pass through the centre c, and it will meet the di- 
rection of the power Q in some point E. Prolong the 
line ME to ES, and through the point E draw the vertical 
er; the three lines eq, er, es being included in the 
same plane, which is that of the wheel, we can decom- 
pose the power Q into two forces R, S, directed along EG, 
EH. For this purpose, we will represent this power by 
the parts ef of its direction ; and completing the paral- 
lelogram EGFH, we will have 

Q : s : : EF : EH, 

Q : R : : EF : eg, or fh ; 

or, since, by drawing the radius CD, the two rectangular 
triangles cde, ehf will be similar, and give 



EF : fh : eh : : ce : cd : de, 



we shall have 



WHEEL AND AXLE. 



Q : s : : CE : de, 
Q : R : : CE : CD. 



167 



Thus, in place of the power Q, we can take the two 
forces R5 S, whose values 



R= 



QXCD 
CE ' 



s= 



QXDE 



CD 



are known, since the direction of the force Q being 
given, all parts of the triangle cde are known. 

Fig, 69. 




Now, of the two forces R, S, the latter being directed 
towards the axis of the cylinder which is immoveable, 
it may be regarded as immediately applied to the point 



li 



158 STATICS. 

C, and destroyed by the resistance of the axis; this 
force, then, cannot contribute in any manner to the 
motion of rotation of the cylinder, and it has no other 
eflfect than that of pressing the pivots against their 
supports. Hence, there remains only the force R which 
can be employed to produce an equilibrium with the re- 
sistance P. 

Now, in the horizontal plane kmen draw the radius 
of the cylinder ki, which will be perpendicular to the 
axis and parallel to me ; also draw the line ke which 
will intersect the axis in some point L. This being 
done, the point l, being in the axis, may be regarded as 
immoveable, and the line ke may be taken for an inflexi- 
ble rod, retained by a fulcrum l, at the extremities of 
which are applied the two forces R, p. Now the direc- 
tions of these two forces are both vertical, and conse- 
quently parallel ; hence in order that equilibrium may 
subsist between them, it is necessary that they should 
be reciprocally proportional to their arms le, lk, or 
that we should have 

R : p : : KL : LE. 

But the similar rectangular triangles kil, lce, give 

KL : LE : : KI : CE ; 

hence we shall have 

R : p : : KI : CE. 



WHEEL AND AXLE. 159 

Hence by multiplying in order, this proportion and 
the following 

Q : R : : CE : CD, 

which we found above, we shall have, in case of equi- 
librium, 

Q : p :: Ki : CD; 

that is to say, the power will he to the resistance as the 
radius of the cylinder is to the radius of the wheel; 
which is the answer to the first question. 

By making the product of the extremes equal to that 
of the means in the above proportion, we have 

QXCD=PXKI. 

The line ce meets the circumference of the wheel at 
the point d'; imagining that there is through this point 
a force q' equal to Q directed along the tangent d'q', we 
shall have (31) 

qxcd=q'xcd'=pxki ; 

from which it appears, that in the case of equilibrium, 
the moments of the power q', equal to Q, and of the re- 
sistance P, both taken with reference to the vertical 
plane passing through the axis of the cylinder, are equal 
to each other. 



160 



STATICS. 



III. 

165. The pressure which the trunnions exert against 
their points of support is evidently the eflfect only of 
the forces p, Q, and of the weight T of the machine, 
which may be considered as united at its centre of 
gravity g^ where, by taking the two forces R, S, instead 
of the power Q, these pressures are produced by the four 
forces p, R, s, t. 

Fig. 69. 




These forces are all known ; for, 1st, the resistance 
p and the weight T of the machine are given imme- 
diately ; 2d, the two other forces K and S, whose values 
we have found to be, in (general. 



QXCD 
CE ' 



QXDE 
CD ^ 



WHEEL AND AXLE. 161 

become, in the case of equilibrium where we have 

QXCD=PXKI, 



PXKI 
CE 



PXKIXDE 

S= , 

CDXCE 



and containing only known quantities, since the direc- 
tion of the power Q being given, we can find all the 
parts of the right-angled triangle cde. / 

Now the two forces p, r, whose directions are vertical, 
and which are in equilibrium about the point L, exert 
upon this point of the axis a weight whose direction is 
vertical, and which is equal to their sum P+R. More- 
over, this weight P+R, being sustained by the two points 
of support at A and B, must be regarded as the resultant 
of the two vertical pressures which it exerts at these 
points; and we will find each of these pressures, by 
dividing the resultant p+r into two parts reciprocally 
proportional to the distances of the point L from the 
two supports. Let x be the pressure exerted at the 
point A, and x' that exerted at the point B ; we will find 
these two pressures by the following proportions : 

AB : BL : : p+r : x, 
AB : AL : : p+r : x'. 

In like manner the weight T of the whole machine, 
supposed to be united at the centre of gravity ^, may 
be regarded as the resultant of the vertical pressures 



162 



STATICS. 



Fig. 69. 




which it produces upon the two points of support : and 
we will find these pressures by dividing the weight T 
into two parts reciprocally proportional to the distances 
Kg, gB. 

Then let y and y' be the pressures which result re- 
spectively at the points A and b; we will find these 
pressures by the two proportions which follow : 

AB : B^ : : T : Yj 
AB : A^ : : T : y'. 



Lastly, the horizontal force S, applied to the point c 
of the axis, produces horizontal pressures upon the two 
supports, directed perpendicularly to the axis AB, and 
of which it is the resultant : we will find likewise these 
pressures by dividing the force S into parts reciprocally 
proportional to the lines AC, cb. Then let z, 7/ be the 
horizontal pressures, produced respectively upon the 



WHEEL AND AXLE. 163 

points Aj B : we will find these pressures by the pro- 
portions : 

AB : BC : : s : z, 
AB : AC : : s : z'. 

Thus the point of support A sustains the two vertical 
pressures x, Y, and the horizontal pressure which acts 
in the direction of the force S. In like manner the point 
B sustains the vertical pressures x', y', and the horizon- 
tal pressure z'. Hence, by compounding, for each of 
these points, the forces which act upon it, we will find 
the intensity and direction of their resultant ; and we 
shall have the intensity of the resistance of which it 
must be capable, as well as the direction in which it 
must resist, so as not to yield to the united efibrts of the 
resistance p, the power Q, and the weight T of the ma- 
chine ; which is the object of the second question. 



IV. 



166. Hitherto we have regarded the cords as infinitely 
fine threads ; but, when the weight P is suspended by 
the cord KP, the line of direction of this weight does 
not coincide with the axis of the cord ; and in the case 
where the cord by wrapping around the cylinder does 
not change in figure, its axis is always at a distance 
^from the surface of the cylinder equal to the semi-di- 
ameter of the cord. Thus, by reason of its thickness, 
the cord is in the same condition as though, being in- 
finitely fine and reduced to its axis, it were wrapped 



164 



STATICS. 



upon a cylinder whose radius was greater than the radius 
of the cylinder of the machine, by a quantity equal to 
the semi-diameter of the cord. It is the same with the 
cord of the wheel, which, by reason of its thickness, 
may be regarded as a mathematical line wrapped upon 
a wheel whose radius is greater than that of the wheel 
of the machine, by a quantity equal to the semi-diameter 
of this cord. Hence, in all the relations we have just 
found, it is necessary to augment the radius of the 
cylinder and that of the wheel by quantities respectively 
equal to the semi-diameters of the cords which envelope 
them. Thus, for example, in the case of equilibrium of 
the wheel and axle, the power Q is to the resistance p as 
the radius of the cylinder ^ augmented hy the radius of 
the cord kp, is to the radius of the wheel^ augmented ly 
the radius of the cord dq. 



Fig. 70. 




167. If several wheels and axles 
are so arranged that the cord bq of 
the first wheel, being drawn by a 
power Q, the cord CE of the cylin- 
der of this wheel, instead of being at- 
tached immediately to the resistance, 
is wound around the second wheel ; 
the cord fh of the cylinder of the 
second is likewise wound around the 
wheel of the third ; and lastly, the cord 
IP of the cylinder of the last is applied 
to the resistance p ; the power and the 



WHEEL AND AXLE. 166 

resistance Tvill not be in equilibrium, unless there is also 
equilibrium between the two forces which act upon each 
separate wheel. Thus, by naming K the tension of the 
cord CE, and L that of the cord FH, in the case of 
general equilibrium we have : Ist, by reason of the 
equilibrium of the first wheel and axle (164), 

Q : K : : CA : AB ; 
2d5 by reason of the equilibrium of the second, 

K : L : : DF : DE ; 
3d, by reason of the equilibrium of the third, 

L : p : : ai : GH ; 

and so on, whatever may be the number of machines. 
Hence, by multiplying all these proportions in order, we 
have 

Q : p : : caxdfxgi : abxdexgh ; 

that is to say, the power is to the resistance as the pro- 
duct of the radii of the cylinders is to the product of 
the radii of the wheels. 

For example, if the radius of the wheel of each is 
four times the radius of its cylinder, in the case of 
equilibrium of all three wheels and axles, we have 

Q : p : : 1x1x1 : 4x4x4, or : : 1 : 64. 



166 



STATICS. 



Hence we see, that by multiplying in this manner the 
number of machines we can put moderate powers in a 
state of equilibrium with very great resistances ; but 
this arrangement is almost never employed, because it 
requires too great a length of cord in the first wheels, 
while the space which the resistance has to traverse is 
inconsiderable. 



VL 



^^- 74. 168. When we wish to profit by 

the advantap;es of this arranore- 
ment, 1st, the cords are suppressed 
which transmit the motion from one 
wheel and axle to another ; 2d, on 
the circumference of each wheel, 
teeth are placed equally distant 
from each other; 3d, on the ar- 
bor of each of the toothed-wheels^ 
another wheel similarly toothed, 
of a smaller diameter, and which is 
called the pinion, is fixed solidly ; 4th, lastly, the 
whole system is so arranged that the teeth of each 
pinion interlock with the teeth of the following wheel. 
By this means one wheel cannot turn upon its axis, with- 
out its pinion communicating motion to the wheel with 
which its teeth are engaged, and causing it to turn upon 
its axis ; and the number of revolutions which each 
pinion can cause the following wheel to make, is un- 
limited. The system is then in the same condition as 
though the pinion, considered as the cylinder of a wheel 




COG WHEELS. 167 

and axle, and the ^'heel with which it interlocks, were 
embraced by the same cord as in the preceding case. 

Hence, when a power Q, applied to the circumference 
of the first wheel, is in equilibrium with a resistance P, 
applied to the circumference of the last pinion, the 
power is to the resistance as the product of the radii 
of the pinions is to the product of the radii of the 
wheels. 



VII. 



169. Toothed wheels are employed in a very great 
number of machines, principally in mills and in the 
works of time pieces. Their immediate object is to 
communicate to a cylinder or arhor a movement of ro- 
tation about its axis, by the aid of the rotary motion of 
another arbor. For this purpose it is not necessary that 
the axes of the two arbors should be parallel, as we 
have heretofore supposed ; it is sufficient that they are 
in the sam6 plane. 

170. When the axes of the two arbors 
are at right angles, the teeth ordinarily 
are placed perpendicularly to the plane 
of the wheel, as may be seen in Fig. 72 ; 
then they can interlock with those of the 
pinion, or with the staves of the trundle 
AB, which produces the efi'ect of a pinion : 
by this interlocking, the teeth of the 
wheel are forced to slide upon the staves 
in the direction of the axis of the trundle, which also 
causes friction. 




168 



STATICS. 



^^^' "^^^ 171. In general, whatever 

may be the angle bag, which the 
axes of the two arbors make, 
so that the movement of rota- 
tion of the one is communicated 
to the other by means of two 
toothed wheels DEFa, ehif, and 
the teeth do not slide upon each 
other in the direction of the axes, it is necessary that 
these two wheels, called beveled wheels^ should be trunca- 
ted sections of two cones dae, eah, which have the 
same summit A, and whose axes coincide with those of 
the arbors ; moreover, the teeth of the two wheels should 
be terminated by conic surfaces which have for their 
summit the common point a. 




VIII. 

172. The jacJc'Screw is also a machine which may be 
referred to the wheel and axle. 

Fig.l^. The simple jack-screw is composed of 

a bar of iron ab, furnished with teeth 
upon one of its sides, and moveable in 
the direction of its length within a box 
DE. The teeth of the bar interlock with 
those of a pinion c, which is turned 
upon an axis by means of a crank F. 
The teeth of the pinion carry along those 
of the bar, and cause the weight to rise, which rests 
upon the head A of the bar, or which is raised by the 
hook B. This machine, evidently, is nothing else than 
a wheel and axle, and, in the case of equilibrium, by 




THE INCLINED PLANE. 



169 



supposing the direction of the power to be perpendicu- 
lar to the arm of the crank, the power is to the resist- 
ance as the radius of the pinion is to the arm of the 
crank. 

In the compound jack-screw, the teeth of the first 
pinion interlock with those of a second toothed wheel, 
and the teeth of the pinion of this wheel interlock with 
those of the bar. By this means the power is placed in 
a state of equilibrium with a greater resistance. This 
case is related to that of toothed wheels, and we will 
not dilate any more upon that subject. 

173. When two powers are in equilibrium by means 
of a pulley or wheel and axle, it is very evident that 
we may consider them as though they were in equili- 
brium at the extremities of a lever whose fulcrum is in 
the axis of the cylinder or the pulley ; hence (151), 
these powers are to each other reciprocally, as the 
spaces which they would traverse along their directions, 
if the equilibrium were disturbed infinitesimally. 



Article III. 



On the Equilibrium of the Inclined Plane, 




174. A plane is said to be inclined, 
when it makes an angle with a horizon- 
tal plane abee, and this angle is not a 
right angle. 

175. If through any point H, taken 
upon the line of intersection ab of 
the inclined plane and the horizontal 



170 STATICS. 

plane, two perpendiculars be drawn to this line, the one 
HK in the horizontal plane, and the other hi in the in- 
clined plane, the angle ihk, formed by these two per- 
pendiculars, is the measure of the inclination of the 
plane. The plane of the angle ihk, which passes 
thK)ugh two lines perpendicular to ab, is perpendicular 
to the line ab; hence it is perpendicular to the two 
planes abcd and abef, which intersect in this line; 
hence this plane is at the same time vertical and per- 
pendicular to the inclined plane. 

176. Reciprocally, every plane which is at the same 
time vertical and perpendicular to the inclined plane, is 
perpendicular to the intersection ab of the inclined 
plane with the horizontal plane ; hence, the lines hi and 
HK along which it intersects these two last planes, are 
also perpendicular to ab, and form between them an 
angle ihk, which is the measure of the inclination of the 
plane abcd with reference to the horizontal plane. 

177. If through the point i there be drawn a vertical 
line IL, this line will not leave the plane of the angle 
IHK ; it will meet the horizontal line hk to which it will 
be perpendicular, and it will form a rectangular triangle 
IHL. The hypotenuse of this triangle is named the 
length of the inclined plane ; the side IL is its height^ 
and- the other side hl is its base. 



178. When a body, which, in a single point Q, touches 
an immoveable plane abcd, is pushed by a single force 
p, whose direction pq, 1st, passes through the point of 



THE INCLINED PLANE. 



171 




Fig- 76. contact Q, 2d5 is perpendicular 

to the plane, this body remains 
at rest. 

Thus, we can conceive the 
force P to be applied to the point 
Q of its direction. Now this di- 
rection being perpendicular to 
the plane, and consequently to 
all the lines QR, QS, qt, which we can draw in the plane 
through the point Q, it is similarly disposed with 
reference to all these lines ; there is then no reason why 
the point Q should move along one of the lines rather 
than along any other; hence this point, and conse- 
quently the whole body, will remain at rest. 

Fig- 77. 179. Both the conditions just men- 

tioned are necessary to the rest of the 
body: for, 1st, if the direction pi 
of the force does not pass through 
the point of contact, nothing prevents 
the point a of the body which is upon 
this direction from approaching the 
plane, and the body would be put in motion. 2d. If 
the direction PQ of the force passes through the point 
of contact, and is not perpendicular 
to the plane, by conceiving this force 
to be still applied to the point Q, let 
its direction be prolonged to QR, and 
through the point Q draw the line QS 
perpendicular to the plane ; through 
the tvro lines qr, qs draw a plane, 
which will intersect the first plane 
ABCD somewhere in a line hi, which will pass through 





172 



STATICS. 



the point Q. This being done, if we represent the force 
P by the part QE of its direction, and complete the pa- 
rallelogram QTRS, instead of the force P we may take 
the tAYO forces represented by QS and qt. Now the 
force QS, which passes through the point of contact, and 
whose direction is perpendicular to the plane abcd, will 
be destroyed by the resistance of this plane (178) ; but 
nothing will oppose the action of the force qt, whose 
direction is parallel to the plane ; the point Q will then 
move in the direction qh, and the body will not be at 
rest. 

180. Hence it follows, that, when a 
body, affected by the simple action 
of its gravity, remains in equilibrium 
upon a plane abcd, which it touches 
in a single point Q, 1st, the centre 
of gravity p of this body, and the 
point of contact Q, are in the same 
vertical line ; 2d, the plane abcd is 
horizontal ; for, since the weight of the body can be 
regarded as a single force applied to its centre of gravity, 
the body cannot be at rest, unless the direction of this 
force passes through the point of contact and is perpen- 
dicular to the plane upon which the body is sustained. 
Fig. 79. 181. It follows again, that 

when a body, affected by the 
single action of gravitation, is 
sustained upon an inclined plane 
ABCD by a single point Q, and 
this point is found in the verti- 
cal drawn through the centre of gravity, this body must 
tend to slip upon the plane ; and the direction QH along 





THE INCLINED PLANE. 



173 



Fig, 80. 




which the point Q tends to move, is the intersection of 
the plane abcd with the plane ihk, which is at the 
same time vertical and perpendicular to the inclined 
plane. 

182. What has just 
been said of a body 
pushed by a single force 
against a plane, must ap- 
ply to a body pushed 
against a curved surface 
AQB, which it touches only 
in the single point Q : that 
is to say, this body cannot 
be at rest, unless the direction of the force which pushes 
it passes through the point Q, and is perpendicular to 
the curved surface at this point ; for this body may be 
considered as sustained upon the plane de tangent to 
the curved surface at the point Q. 

183. We see, then, that when a lever may slide upon 
its fulcrum, it is not sufficient for its remaining at rest, 
that the resultant of the tvfo powers which are applied 
to this lever are directed towards the fulcrum ; it is be- 
sides necessary that the direction of this resultant 
should be perpendicular to the surface of the lever at 
the point where it touches the fulcrum. 



II. 

184. When a body, pushed by a single force p 
against an immoveable plane abcd, is sustained upon 
this plane by a definite base vxyz, if the direction PQ 



174 



STATICS. 



Fig. 81. 




of the force meets the base somewhere in a point Q, and 
if it is at the same time perpendicular to the plane, the 
body is at rest ; for we have seen (178) that if the base 

w^ere reduced to the single 
point Q, rest would take place ; 
it is evident that the other 
points of support, which the 
base presents, cannot disturb 
it. 



We will demonstrate, as in 



No. 179, that both these con- 
ditions are necessary to the 
body's remaining at rest : the effect of the first is to 
prevent the body from turning upon one of the sides of 
its base ; the effect of the second is to prevent it from 
slipping upon the plane abcd. 

185. If the body, instead of being supported upon 
the plane by a continuous base, simply touches it by 
several points separated from each other, we may regard 
these points as the summits of the angles of a polygonal 
base ; and the body is at rest when the direction of the 
force, which pushes it against the plane, is perpendicu- 
lar to this plane, and at the same time passes through 
the interior of the polygon. 

Thus, since the weight of a body may be regarded as 
a vertical force applied to its centre of gravity p, we 
see, 1st, that a body, which rests with its base upon a 
horizontal plane abcd, cannot be stable, unless the ver- 
tical PQ, drawn through the Centre of gravity, meets 
some point Q of the base ; 2d, if the body rests upon 
the plane by a certain number of points of support. 



THE INCLINED PLANE. 



Fig. 82. 




u, x, Y, . . . it cannot be stable, 
unless the vertical pq, drawn 
through its centre of gravity 
F, passes through a point 
Q taken in the interior of the 
polygon UXY, vfhicli may be 
formed by joining the exterior 
points of support by the lines 
ux, XY, YU. 



III. 



186. Hitherto we have supposed that the body, sup- 
ported upon a plane, was pushed by a single force ; but it 
is evident that, if the body is pushed by several forces 
at the same time, it cannot be at rest, unless the result- 
ant of all these forces satisfies the preceding conditions ; 
that is to say, unless the direction of this resultant is 
perpendicular to the plane, and passes through the base 
of the body. There is then in this case a third con- 
dition necessary to the rest of the body, and this condi- 
tion is, that all the forces which act upon it must have 
one resultant. 

The whole theory of the equilibrium of a body, pushed 
by as many forces as we please to assume, and supported 
upon a single resisting plane, consists in the search for 
the directions and intensities which the forces must have, 
so that the three conditions just mentioned may be ful- 
filled. We will content ourselves with developing it for 
a few simple cases, and principally for that in which the 
body is pushed by two forces. 



176 



STATICS. 



IV. 



Fig. 81. 




187. Let Luz be a body sup- 
ported by a base UXYZ upon a 
resisting plane abcd, and so- 
licited at the same time by two 
forces R, s. According to the 
preceding remarks, in order 
that the body may be at rest, 
it is necessary, 1st, that the 
two forces R, s should have 
one resultant : now, two forces cannot have one result- 
ant, unless their directions are in the same plane, (10) ; 
hence, 1st, the directions of the two forces R, s must be 
included in the same plane, and intersect in a certain 
point N. 

2d. It is necessary that the direction pnq of the re- 
sultant of the two forces R, S, should be perpendicular 
to the plane abcb : now^ the resultant of the two forces 
is always contained in the plane drawn through their 
directions ; hence, 2d, the plane in which the two forces 
R, S are directed, should be perpendicular to the plane 

ABCD. 

3d. It is necessary that the direction of the resultant 
should pass through a point Q of the base. 

188. From this it follows, that if one of the two 
forces, for example the force R, is the weight of the 
body which may be considered as applied to the centre 
of gravity M, and whose direction MR is vertical, the 
body cannot be at rest upon an inclined plane abcd, 
unless the direction NS of the other force is contained 



THE INCLINED PLANE. 



177 



in a vertical plane, drawn through the centre of gravity 
M, perpendicularly to the inclined plane, and, moreover, 
that the direction PQ of the resultant of the two forces 
is perpendicular to the inclined plane, and passes through 
a point Q of the base of the body. 

Now all these conditions, relative to the directions of 
the forces, being supposed to be filled, we will seek the 
relations which the two forces E, S, and the weight P 
of the plane, have for each other in the case of equi- 
librium. 




V. 

^^9' 83. 189. Let Lxu be a body sup- 

ported by its base ux upon a re- 
sisting plane hi, and kept at rest 
upon this plane by the two forces 
R, s. Having prolonged the di- 
rections of the two forces until 
they meet in a point N, draw 
R. through this point the line NP per- 

pendicular to the plane hi. We have seen that this 
line will be the direction of the resultant of the two 
forces R, S. Hence, if we represent this resultant by 
the part NE of its direction, and if, by drawing through 
the point E the lines Ea, ef, parallel to the directions 
of the forces e, s, we complete the parallelogram nfeg, 
the sides nf, Na will represent the intensities of the 
forces E, S. Hence, by naming p the weight on the 
plane which is equal to the resultant, we shall have 



R : s : p : : NF : Na or FE : NE. 



178 STATICS. 

In order to have the ratios of these three forces ex- 
pressed in quantities independent of the construction 
of the parallelogram NFEa, we will remark, that in the 
triangle nef the sides are to each other in the ratio of 
the sines of the opposite angles, or we shall have 

NF : FE : EN : : sin nef : sin fne : sin nfe. 

Hence we shall have 

R : S : P : : sin nef : sin fne : sin nfe. 

Now these three angles are those which form with each 
other the directions of the three forces E, S, P ; hence 
these forces are to each other as the sine of the angle 
which the directions of the other two forms. 

We see then that, of the six t^ngs 
which may be considered in this 
equilibrium, and which are, the di- 
rections of the three forces and their 
intensities, any three being given, we 
can determine the other three, in all 
cases where, of the six things which 
we consider in the triangle nef, namely, the angles and 
the sides, the three analogous ones being given, we can 
determine the three others. 

190. If one of the forces, for example the force R, 
is the weight of the body whose direction is vertical, 
and passes through the centre of gravity M, and the di- 
rection of the force S, which retains the body in equi- 
librium upon the inclined plane, is parallel to this plane, 
draw the base kh and the height ki of the inclined 




THE INCLINED PLANE. 



179 



plane ; the right-angled triangles nfe, ihk will be simi- 
lar, because the angles nee, hik, whose sides are paral- 
lel each to each, will be equal, and we shall have 

* KF : fe : en : : HI : IK : KH ; 

hence we shall also have 

R : S : P : : HI : IK : kh. 

Thus, in this case, the weight of the body is to the 
force which holds it in equilibrium, as the length of the 
inclined plane is to its height. 

Fig- 85. _ 191. By supposing still that 

the force R is the weight of the 
body, if the direction of the 
force S is horizontal, and con- 
sequently parallel to the base 
hk of the inclined plane, the 
right-angled triangles nfe, hki are again similar ; be- 
cause the sides of the one will be perpendicular to the 
sides of the other, and we shall have 

NF : FE : EN : : HK : Ki : ih, 

we shall also have 




R : s : p : : HK : KI : IH. 



Hence, the weight of the body is to the force which 
holds it in equilibrium, as the base of the inclined plane 
is to its height. 



180 



STATICS. 



VI. 




192. Let us now consider 
the equilibrium of a body sup- 
ported by two inclined planes. 
Let M be a body subjected to the 
^D single action of gravity, and re- 
tained by the two inclined planes 
ABCD, ABEF, wliich intersect some- 
where in the line ab. Let ii, i be the points at which 
the body touches the two planes, and gr the vertical 
line drawn through its centre of gravity, and w^hich will 
consequently be the direction of its w^eight. It is evi- 
dent that this body cannot be at rest, unless its weight 
R can be decomposed into two other forces P, Q, applied 
to the same body, and vfhich are destroyed by the re- 
sistance of the two planes : or, what is the same thing, 
unless the directions of the two forces p, q pass through 
the points of support i, ii, and are perpendicular each 
to the corresponding inclined plane. Now the direction 
of a force and those of its two components are contained 
in the same plane, and necessarily meet in the same 
point ; hence, in order that the body M may be at rest 
between the two inclined planes, it is necessary that the 
perpendiculars IG, HG, drawn through the points of sup- 
port I, II to the two inclined planes, should be in the 
same plane w^ith the vertical drawn through the centre 
of gravity of the body, and intersect this vertical in the 
same point G. 



THE INCLINED PLANE. 181 

193. Hence it follovrs, that, in order that a body m 
may be at rest between two inclined planes, inde- 
pendently of the position of the body, the planes should 
satisfy the condition that the line AB of their intersec- 
tion should be horizontal. Thus, the plane IGH, which 
must contain the vertical or, and the perpendiculars 
la, HG to the two inclined planes, is at the same time 
vertical and perpendicular to these two planes ; hence^ 
reciprocally, the two inclined planes must be perpen- 
dicular to the vertical plane igh ; hence the line ab 
of their intersection should be perpendicular to this 
same plane, and consequently horizontal. 

^^i^'-ST. 194. These conditions, 

y^ ^. which have for object the 

I 9 Wk respective positions of the 

\j(/' ^^^\/'^^^r^y^ clined planes, being sup- 
B?w__\^_j^^^ posed to be fulfilled, in 

\v y^ order to find the ratio of 

^ the weight R of the body 

to the weights P, Q, which the two planes support, we 
will remark that the plane IGH, vertical and perpendicu- 
lar to the two inclined planes containing the angles 
which these two planes form with the horizon, includes 
all that relates to the question, and we may be content 
to consider it alone, as in ¥ig. 87. Hence, let AD, af 
be the intersections of the vertical plane lan with the 
two inclined planes ; these lines form wdth the horizon- 
tal line Vz, or with any other horizontal line XY, angles 
which will measure the inclinations of the two planes. 
This being granted, if we represent the w^eight of the 

16 



182 STATICS. 

body by the part GR of its direction, and complete the 
parallelogram gpqr, we shall have 



E : p : Q : : an : RQ : QG. 

Now the triangles gqr, xay, whose sides are perpendicu- 
lar each to each, give 

GR : RQ : QG : : XY : YA : AX ; 
hence we shall also have 

R : p : Q : : XY : YA : AX ; 

or, lastly, because the sides of the triangle xay are pro- 
portional to the sines of the opposite angles, we shall 
have 

R : p : Q : : sin yax : sin axy : sin xya, 

that is to say, by representing the weight of the body 
by the sine of the angle which the two inclined planes 
form with each other, the weights which these two 
planes support are to each other reciprocally as the 
sines of the angles which these planes form with the 
horizon. 



THE INCLINED PLANE. 



183 



VII. 



Fig. 88. 



,..' — S\ 





195. Lastly, if a body is 
sustained by three points 
A, B, c, upon three inclined 
planes, it is evident that 
this body cannot be at rest 
unless its weight P, whose 
direction dp is vertical, and 
passes through the centre of gravity of the body, can 
be decomposed into three other forces Q, R, S, which are 
destroyed by the resistances of the inclined planes^ 
that is to say, unless the directions of the three forces 
Q, R, S pass through the three points of support, and 
are each perpendicular to the corresponding inclined 
plane. 

196. In order that the body rnay be at rest, it is 
necessary that the weight p should be susceptible of 
being decomposed into two forces Q, x; the first of 
which, Q, being directed towards one of the points of 
support A, perpendicularly to the inclined plane which 
passes through this point, the other force x may itself 
be decomposed into two others R, S, in the directions 
of the two other points of support b, c, perpendicular 
to the other inclined planes. 

Hence we see, in this case, it is not necessary that 
the perpendiculars to the inclined planes, drawn through 
the points of contact A, B, c, should all three meet in 
the same point, nor even that they should all three meet 
the vertical drawn through the centre of gravity of the 
body. 



184 STATICS. 



THEOREM. 



197. When a hody without gravity^ supported upon 
an inclined plane BC hy a single point c, is in equili- 
hrium between two powers P, Q, these powers are to each 
other reciprocally as the spaces which they would tra- 
verse in the line of their directions^ if the equilibrium 
were disturbed infinitesimally . 

^^3' 89. Demonstration. The two 

powers P, Q being in equili- 
brium, their resultant should 
be perpendicular to the inclined 
plane, and pass through the 
point of support c (186) ; it 
follows from this that the di- 
rections of these powers should 
coincide in a certain point a, and that the line go 
should be perpendicular to the inclined plane. This 
being established, from the point c draw upon the di- 
rections of the powers P, Q, the perpendiculars ce, cf : 
it is evident that the angle ecf, supposed to be invaria- 
ble, may be considered as a bent lever, at the extremi- 
ties of which are applied the two powers in equilibrium 
about the point of support c : hence we can demonstrate, 
as in No. 151, that the powers are to each other recipro- 
cally as the spaces which they would traverse in the line 
of their directions, if the equilibrium were disturbed 
infinitesimally. 




THE SCREW. 



185 




On the Screw. 

198. If we conceive a cylin- 
der ABCD to be enveloped by 
a thread aghik . . . ., and so 
disposed that the angles FLO, 
r FMP, FNQ ... .5 formed by 
^' the direction of the thread 
J, with the lines drawn upon the 
surface of the cylinder paral- 
lel to the axis, are equal to each other, the curve which 
the thread traces upon the surface of the cylinder is 
named a lielix, 

199. Hence it follows, that if we develope the surface 
of the cylinder, and extend it upon a plane, as we see 
in ahcd^ 1st, the developement all' or lik! of one revolu- 
tion of the helix will be a straight line ; because the 
angles which this line will form with all such lines as ^/, 
drawn parallel to the side ad^ will be equal to each 
other. 2d. This developement ali' of a revolution of 
the helix will be the hypotenuse of a right-angled tri- 
angle aVh'^ whose base ab will be equal to the circum- 
ference of the base of the cylinder, and whose height 
hli' will be equal to the distance of the revolution which 
we consider^ from the revolution which follows it. 3d. 
All the hypotenuses ali\ hV being parallel to each 
other, the right-angled triangles ahli'^ JiK'k! . . ., will be 
equal and similar, and will have equal heights. Hence 
the intervals lm, mn, . . . between two consecutive revo- 
lutions of the helix, considered upon the surface of the 
cylinder, are everywhere equal to each other, 

IG* 



186 



STATICS. 



Fig. 91. 




Fig. 92. 






1 II 






200. This being established, the screw may be con- 
sidered as a right cylinder, enveloped by a projecting 
fillet, adhering and wound as a helix upon the surface 
of the cylinder. In the wooden screw, the form of the 
fillet is such, that if it be cut by a plane drawn through 
the axis of the cylinder, its section is most frequently 
an isosceles triangle, as may be seen in I^ig, 92 ; but 
in large iron screws which are made with care, the sec- 
tion of the fillet is rectangular, as in Fig, 91. The 
constant interval ab, which is found between two con- 
secutive revolutions of the fillet, is named tlie pitch of 
the screw^ or the helical interval. 

201. The piece mn, into which the screw enters, is 
named the nut. Its cavity is invested with another pro- 
jecting fillet, adhering, and wound likewise as a helix ; 
and whose figure is such that it exactly fills the inter- 
vals left by the fillets of the screw. Thus, the screw 
can turn in its nut, but it cannot do so without moving 
in the direction of its axis ; and for one entire revolu- 
tion, it moves in the direction of the axis by a quantity 
equal to the pitch of the screw. 



THE SCREW. 187 

202. Sometimes the screw is fixed, and the nut 
moves around it; then, for each revolution, the nut 
is carried upon the screw by a quantity equal to the 
interval. 

203. The screw may serve to elevate weights or over- 
come resistances ; but it is employed most generally 
when great pressures are proposed to be exerted. For 
this purpose we apply a power Q to the extremity of a 
bar w^hich traverses the head of the screw, Fig. 92, or the 
nut, FicL 91, according as it is the one or the other of 
these two pieces which is moveable ; and this power, by 
causing the piece to turn to which it is applied, makes 
the head of the screw advance towards the nut, or re- 
ciprocally. The objects to be compressed are ranged 
between two plates ; one of these plates is fixed, the 
other is pressed by the movable piece, which can ad- 
vance only by reducing the volume of the objects. 

We now propose, disregarding the friction, to find the 
ratio of the powder Q to the resistance p, which is in 
equilibrium w^ith it by being opposed to the motion of 
the moveable piece, along a direetioji parallel to the 
axis of the screw ; and, because the effect is absolutely 
the same, whether the screw turns in its nut, or the nut 
turns upon the screw, it will be suflScient to examine the 
latter case. 

204. The screw being fixed and in a vertical position, 
we will conceive the nut to be left to the action of gravi- 
ty, and even, if we please, that it is charged with an 

J- additional weight; it is evident that it will descend by 
turning, and that it will traverse all the interior fillets 
of the screw, by sliding over them as over inclined 
surfaces. It is also evident that w^e oppose this effect 



188 



STATICS. 




^'^3- 91. by preventing the nut from 

turning around the screw, and 
consequently by applying to 
the extremity of the bar rv a 
power Q which is directed per- 
pendicularly to this bar, and 
in a plane perpendicular to 
the axis of the screw. 

205. Suppose, for an in- 
stant, that the nut rests upon 
the surface of the fillet of the screw only in a single 
point ; this point, during the motion of the nut, will de- 
scribe a helix, whose pitch will be the same as that of the 
screw, and which we may conceive to be traced upon 
the surface of a cylinder, whose radius would be equal 
to the distance of the describing point from the axis of 
the screw. Let abcd be the cylinder, efghi the helix 
^^'9' 93. in question, and M the 

point of the nut which 
describes it. Let XY be 
the tangent of the helix 
at the point M ; through 
a point Y of this tangent 
^ draw the vertical YZ, equal 
to the pitch of the helix, and in the vertical plane XYZ of 
the horizontal line xz : it is clear, that this last line will 
be equal to the circumference of the base of the cylin- 
der ABCD (199), or to the circumference of the circle 
whose radius is CK ; and we will represent it by circ. CK. 
This being established, the point M, which we may 
regard as charged with the whole weight P of the screw, 
will be supported upon the helix as though it were upon 




THE SCREW. 189 

the inclined plane xy ; thus, in order to hold it in equi- 
librium by means of a force R, which should be imme- 
diately applied to it, and which should be directed pa- 
rallel to xz, it is necessary (191) that this force R should 
be to the weight P, as the height of the inclined plane 
is to its base, or that we should have, 

R : p : : YZ : circ. ck. 

But if, instead of a force R immediately applied to the 
point M, we employ a force Q, whose direction is parallel 
to that of the first, and which acts at the extremity of 
a bar m'/, it is necessary that this force should exert 
upon the point M the same efi^ect as the force R, and for 
that purpose, these forces should be to each other re- 
ciprocally as their distances from the axis of the cylin- 
der ; that is to say, we should have, 

Q : R : : m/ : mJ; 

or, since the circumferences of circles are to each other 
as their radii, we should have, 

Q : R : : circ. m/ : circ. M J. 

Then, by multiplying together in order this proportion 
and the first, we shall have, since circ. CK=circ. m/, 

Q : p : : YZ : circ. m'/; 

that is to say, the power which retains the nut in equi- 
librium, will be to the weight of the nut, as the pitch of 



190 



STATICS. 



the screw is to the circumference of the circle which the 
power tends to describe."^ 

206. Since the distance of the point M from the axis 
of the screw does not enter into this proportion, it fol- 
lows that, whatever may be this distance, the ratio of 
the weight P of the nut to the power Q, which is in 
equilibrium with it, is always the same, provided this 
power is always applied to the same point. 

207. If the fillet of the nut is supported upon that 
of the screw by several points unequally distant from 
the axis of the screw, as it generally happens, the total 
weight of the nut can be regarded as divided into par- 

F^g- 91- tial weights, each applied to 

one of the points of support. 
Now the partial power ap- 
plied to the point v, and which 
makes an equilibrium with one 
of these partial weights, is to 
this weight, in the constant 
ratio of the pitch of the screw, 
to the circumference which 
the power tends to describe. 
Hence the sum of the partial weights, or the total weight 
of the nut, is to the sum of the partial powers, or to the 
total power Q, in the same ratio. 

208. From this it follows : 1st. The force necessary 
to be applied to the nut parallel to the axis of the screw, 
to produce an equilibrium with the power Q, which tends 




"^ This relation is independent of the form of the fillet of the 
screw ; because the only question is to prevent the moveable piece 
from turning around a line which is supposed to be fixed, No. 203. 



THE SCREW. 191 

to turn the nut, should be to this power, as the circum- 
ference of the circle which this power tends to describe, 
is to the pitch of the screw ; 

2d. For the same screw, the effect of the power Q is 
as much greater as this power is applied at a greater 
distance from the axis of the screw ; 

3d. For two different screws, the power being applied 
at the same distance from the axis, its effect is as much 
more considerable as the height of the pitch is less ; that 
is to say, the closer the fillets of the screw are together, 
the greater is the effect of the power for compressing 
in the direction of the axis. 



II. 

^^•94. 209. The screw is 

sometimes employed for 
^? communicating to a 
toothed wheel a motion 
of rotation upon its 
axis. For this purpose, 
having given to the screw 
a pitch DE, equal to one 
of the divisions of the 
toothed wheel, it is so 
arranged that its axis is 
in the plane of the wheel, and its fillet catches in the 
teeth. This being done, when a power Q turns the screw 
upon its axis, by means of a crank ra, the fillet carries 
along the teeth, which follow each other, and it turns 
the wheel in spite of the resistance P, which opposes 
its rotary motion. 




192 STATICS. 

When the screw is employed for this purpose, it is 
named the endless screw. 

To find the ratio of the power Q to the resistance P 
in the case of equilibrium, suppose the resistance is 
suspended to a weight by a cord which envelopes the 
arbor of the wheel. By virtue of this weight, the tooth 
of the wheel presses the fillet of the screw parallel to 
the axis hf ; and if we name R this pressure, we have 
(164), 

p : R : : AC : AB. 

Now we may regard the pressure of the tooth as that 
which a nut, pushed by a force R parallel to the axis 
of the screw, would exert ; we have, in the case of equi- 
librium, 

R : Q : : circ. ra : de ; 

hence, by multiplying together the two proportions in 
order, we have, 

p : Q : : ACXcirc. fg : abxde ; 

that is to say, the resistance is to the power, as the pro- 
duct of the radius of the wheel by the circumference 
which the crank describes, is to the product of the 
radius of the arbor of the wheel by the pitch of the 
screw. 



THE WEDGE. 



193 



On the Wedge. 




Fig. 95. 210. The wedge is a triangular prism 

ABCDEF, which is introduced by its sharp 
edge EF into a crack, to split or sepa- 
rate the two parts of a body. It is 
also made use of for exerting great 
pressures or to stretch cords. 

Knives, hatchets, punches, and, in 
general, all cutting and penetrating in- 
struments may be considered as wedges. 

The face abcd, upon which we strike the wedge to 
sink it, and which receives the action of the power, is 
named the heel of the wedge; the side ef, by which the 
wedge commences to penetrate, is named the hlade; and 
the name of sides are given to the faces afed, bfec, by 
which it compresses the bodies which it has to separate ; 
or, since we are accustomed to represent the wedge by 
its triangular profile abf, the base ab of the triangle 
is called the heel of the wedge ; af and bf are its sides. 
^^9' 96. 211. We are accustomed to suppose 

that the direction of the power is per- 
pendicular to the heel of the wedge ; 
because, generally, the wedge is sunk by 
striking upon the heel with a hammer, 
or with any other object which has no 
connection with it, and which, in this 
case, if the direction CD of the shock is 
not perpendicular to the surface of the 
heel, the action is naturally decomposed into two other 
forces, CH, ce ; the first of which, being parallel to the 

17 




194 



STATICS. 



heel of the wedge, can have no other effect than that 
of making the hammer slip, and the second being per- 
pendicular to the face ab, is the only one which is trans- 
mitted to the wedge, and conduces to the effect which 
is to be produced. But if the power were applied to 
the wedge by means of a cord, whose point of attach- 
ment could not slip, then, by considering this power, it 
would be necessary to take account of its direction. 



G- < 






^ 

K- 


/-'" 


t\ 


,2v Jf'- 


->^ 


^..^— - 


/m"' 


-t:-..^^ 



^^' 97. 212. Let c, D, be two 

points separated by a 
wedge ABF, and retained 
by a cord CD, which is 
attached to them, and 
which opposes their sepa- 
ration; suppose, more- 
over, that these points 
are supported against a 
resisting plane which 
prevents them from moving in a direction perpendicu- 
lar to the cord, and that a power p, applied perpendicu- 
larly to the heel ab of the wedge, makes an effort to 
separate them and move them in the direction of the 
length of this cord. This being established, it is re- 
quired to determine, 1st, the tension which results in 
the cord CD ; 2d, the force necessary to be applied to 
one of the two points c, d, to prevent them from moving 
in the direction of the cord ; 3d, the pressure which 
each of these points exerts upon the plane which resists 
them. 



THE WEDGE. 195 

It must be remarked first, that, if the direction of the 
power p is not such that it may be decomposed into two 
others Q, R, whose directions pass through the points 
c, D, and are perpendicular to the sides AF, be, the 
wedge will turn between the two points C, D until this 
condition is fulfilled, and then only will the power p 
produce all its efiect. We will suppose, moreover, that, 
having drawn through the points c, D the lines CE, de 
perpendicular to the sides of the wedge, the point of 
intersection e of these two lines is in the direction of 
the power p. 

This being the case, the force p will be decomposed 
into two other forces Q, R, directed along EC, ed ; and, 
if we represent this force by the part ex of its direc- 
tion, and finish the parallelogram ezxy, we shall have, 

p : Q : R : : EX : EY : EZ or YX ; 

or, since the two triangles exy and abe, whose sides are 
perpendicular each to each, are similar, we shall have, 

p : Q : R : : AB : AE : BF, 

and, consequently. 



PXAF PXBF 

Q= , R= . 

AB AB 



The point c not being able to move in the direction 
ECH, because of the resistance of the plane upon which 
it is supported, the force Q, which is applied to it, will 
be decomposed into two others ; one of which, in the 



196 



STATICS. 



direction of the line ci, perpendicular to the cord, will 
be destroyed by the resistance of the plane, and the 
other, in the direction of the prolongation of DC, will 
be employed in stretching the cord. Thus, by making 
CH=EY, and completing the rectangle cani, the two 
components of the force Q will be represented by ci and 
CG ; and we shall have, 

Q : force ci : force ca : : ch : ci : ca, 

and, consequently, 

force ci= — , 



force ca= 



CH 

Qxca 



CH 



or, substituting for Q its value previously found, we shall 



have. 



force ci= 



PXAFXCI 




ABXCH 
PXAFXCG 
ABXCH 

In like manner, if, 
upon the prolongation 
of ED, we make dl=ez, 
and complete the rect- 
angle DKLM, whose side 
DK is upon the prolonga- 
tion of CD, and whose 
side DM is perpendicular 
to CD, the force R will 
be decomposed into two 



THE WEDGE. 197 

others dm, dk, the first of which will be destroyed by 
the resistance of the plane, and the second will be 
wholly employed in acting upon the cord ; and we will 
find, likewise, 



RXDM PXBFXDM 

force DM= = , 

DL ABXDL 

p RXDK PXBFXDK 

force DK= = -. 

DL ABXDL 



Thus the cord CD will be drawn in one direction by the 
force ca and in the contrary direction by the force dk. 

Now, when a cord is drawn in contrary directions by 
two unequal forces, the tension which it sufi*ers is always 
equal to the smaller of these two forces ; for, when the 
two forces are equal, one of them is the measure of the 
tension of the cord ; and when they are unequal, the 
excess of the greater over the smaller, not being coun- 
ter-balanced by anything, does not contribute to stretch 
the cord, and has no other efi*ect than to draw it along 
in the direction of its length. 

Hence, 1st, the tension of the cord cd will be equal 
to the smallest of the two forces ca, dk. 

2d. The cord will be drawn along in the direction of 
its length and in that of the greater of the two forces 
ca, DK ; so that, in order to oppose this motion, it will 
be necessary to apply to one of the two points c, d a 
force equal to the difference of these two forces, and 
directly opposed to the greater. 

8d. The pressures exerted by the two points c, D 
upon the plain which retains them, will be equal ; the 
first to the force ci, the second to the force dm. 



198 



STATICS. 




IL 

213. If the sides af, bf 
of the wedge be equal, the 
heel AB is parallel to the 
cord which retains the two 
points c, D ; and, at the same 
time, if the direction of the 
^R force P is perpendicular to 
the middle of ab, 1st, the 
wedge will not turn, because the lines ce, de, drawn 
through the two points of support perpendicular to the 
sides of the wedge, will meet in a point E of the di- 
rection of the power. 2d. Both sides being perfectly 
alike, the forces ca, dk will be equal to each other, and 
each of them will be the measure of the tension of the 
cord CD. 3d. By letting fall from the point F the per- 
pendicular FN upon the heel of the wedge, the two tri- 
angles CGH, BNF, whose sides will be perpendicular each 
to each, will be similar, and give, 



CH : ca : : BF : FN. 
Now we have, 

Q : force cg : : ch : ca, 

and, consequently, 

Q : force ca : : bf : FN ; 

But, wc have, also, 

p : Q : : A?. : BF. 



THE WEDGE. 



199 



Hence, by multiplying together these two proportions 
in order, we shall have 

p : force CG : : ab : FN ; 

that is to say, in this case, the power P will be to the 
tension of the cord CD, as the heel of the wedge is to 
its height. 

We will not make any application of the theory of 
the wedge to the use which may be made of this in- 
strument for splitting bodies ; because, under such cir- 
cumstances, the resistance which has to be overcome is 
always unknown, and it is useless to know the ratio of 
this resistance to the power which is in equilibrium 
with it. 



LEMMA. 

214. If from the summits b, c of two of the angles 
of a triangle^ the perpendiculars be, cd he dropped upon 
the opposite sideSy these perpendiculars will be to each 
other reciprocally as the sides upon which they are 
dropped ; that is to say^ we shall have^ 



be : CD : : a:b : ac. 



Fig. 99. 




Demonstration. By con- 
sidering AB as the base of 
the triangle, the perpen- 
dicular CD will be its height, 
and the surface of the tri- 

ABXCD 



angle will be 



In 



200 STATICS. 

like manner^ by taking AC for the base, the surface will 
be — ^ — ; hence, we shall have the two equal products 
ACXBE=ABXCD5 which wiU givc the proportion, 

BE : CD : : AB : AC. 



THEOREM. 

215. When the two powers P, Q are applied to the 
faces AB, AC of a wedge, the thij^d face of which BC is 
supported upon a resisting "plane mn", and these two 
powers are in equilihriur)i through the resistance of the 
plane, they are to each other reciprocally as the spaces 
which tJiey traverse in the line of their directions, if the 
equilibrium he disturbed infinitesimally . 

Fig. 100. 



Demonstration. Since the powers p, q are in equi- 
librium, their directions are perpendicular to the faces 
of the wedge to which they are applied (179). Now 
suppose, by virtue of a derangement in the equilibrium, 
the wedge slips upon the resisting plane, and takes the 
infinitely near position ahc ; and the direction QE is pro- 
longed until it meets ac in e: it is evident that the 



THE WEDGE. 201 

small lines E^, Bd will be the spaces which the powers 
will have traversed in the lines of their directions. 

Lastly, draw Aa ; prolong CA, ha until they intersect 
somewhere in F, and from the points A, a let fall the 
perpendiculars ah, aa ; we will have evidently Ee=Ga 
and Dd=AB.. 

This being determined, the powers p, q being in equi- 
librium, they are to each other as the sides of the 
wedge to which they are applied ; then we shall have 
(212), 

p : Q : : AE : AC, 
and, because of the similar triangles abc, FaA, 

p : Q : : Fa : FA ; 

hence (214), we shall have, 

p : Q : : aa : AH 
and, consequently, 

p : Q : : Eg : T>d. 

216. We have seen that the analogous proposition 
takes place in the equilibrium of all the machines which 
we have considered. By following the same process, it 
could be demonstrated directly that, when two powers 
are in equilibrium by means of points of support which 
any machine presents, they are to each other reciprocally 
as the spaces which they would traverse in the lines of 
their directions, if the equilibrium were infinitesin ally 



202 STATICS. 

disturbed. By means of this proposition, it will be 
easy to find in practice the relation which should subsist 
between a power and a resistance applied to a proposed 
machine, in order that these forces should be in equi- 
librium, leaving out of consideration friction and other 
obstacles to motion. 



203 



NOTE. 



A NEW DEMONSTRATION OF THE PARAL- 
LELOGRAM OF FORCES. 

BY M. CAUCHY.* 

If we suppose, as in general, a force to be repre- 
sented by a length laid off from its point of application 
along the direction in which it acts, the resistance R of 
two forces p, q, simultaneously applied to a material 
point (a), will be represented in intensity and direction 
by the diagonal of the parallelogram constructed upon 
these two forces. This proposition has already been 
demonstrated in several manners. But, among the 
demonstrations which have been given, some require the 
consideration of new material points connected with the 
point (a) by rigid and invariable lines ; others are 
founded upon the use of the differential calculus, or of 
derived functions; others again are deduced from the 
relations which exist between the cosines of certain 
angles. I am here about to demonstrate the same propo- 
sition without recurring to these different considerations, 

•^ This demonstration is extracted from the work which M. Cauchy 
publishes in numbers under the name of Exercises de 3Iathematiques. 



204 STATICS. 

and, in order to attain my object, I will establish suc- 
cessively several lemmas, which may be enunciated as 
follows. 



LEMMA I. 

If we designate hy R the resultant of the two forces 
P, Q, simultaneously applied to the point (a), and by x 
any numler^ the resultant of two forces^ equal to the pro- 
ducts P:r, Q^, and directed along the same lines as the 
forces P, Q, will he represented hy the product ^x^ and 
directed along the same line as the force R. 

fvyt 

Demonstration. In the first place let x^—. m and 

n 

n representing any two whole numbers. We shall have 



p 


Q 


Fx=7n-. 


Qx=m-, 


n 


n 



p 

Besides, we may consider the component p, or m -, as 

p 
produced by the addition of several forces equal to -, 

and the component Q, or m - , as produced by the addi- 
tion of as many forces equal to -. From this it is easy 

to conclude, that the resultant of the forces vx^ Qx, will 

both be produced by the addition of several forces 

P Q ... 

equal to the resultant of - and -. Moreover, it is evi- 
^ ' n n 

dent that the first two resultants are to the last, as the 



NOTE. 205 

numbers n and m are to unity. Hence, the second re- 
sultant will be equivalent to the first multiplied by the 

ratio —=x^ that is to say, to the product ^x. 
n 

Let us suppose in the second place that the number 

X is irrational. Then we can vary the whole numbers 

m and n in such a way that the fraction — will converge 

n 

to the limit x ; and it is evident that, in this case, the 

resultant of the forces — , — , always directed alons; 

n n 

the same line, and always equal to — , will tend more 

and more to coincide, in intensity and direction, with the 
resultant of the forces vx^ (^x. Hence, this last result- 
ant will be directed along the same line as the force R, 
and it will have for measure the limit of the product 

— , that is to say, the product ^x. 



Corollary. 

A 
If we designate by the notation (p, q) the angle in- 
cluded between the directions of the two forces P and Q 

A A . 
(p, r), (q, r) will be the angles included between the di- 
rections of the components P and Q and their resultant 
R. This being fixed, if we make successively Ra:=P, 

P Q 

Ra;=Q, or, what amounts to the same, x=-. a;=-, we 

. R r' 

18 



206 STATICS. 

■will conclude from the preceding lemma, Ist, that the 

p2 pQ 

force P may be replaced by two components — and — , 

II II 

A A 

which form with it angles equal to (p, r) and (q, r) ; 
2d, that the force Q may be replaced by two com- 

A 



^2 



Q PQ 

ponents — and — , which form with it the angles (q, r) 

K R 

A 
and (p, r). 



LEMMA II. 

The resultant R of two forces p, Q, which intersect at 
right angles^ is represented in intensity ly the diagonal 
of the triangle constructed upon the tivo components^ so 
that we have^ 

R2 = p2 + Q2 (1) 

Demonstration. Let us conceive the force p to be 
replaced by the two above-mentioned components, that 

p2 PQ 

is to say, by two forces — and — , which form with 

R R 

A A . . , 

it the angles (p, r) and (q, r). Let us conceive also that 

q2 PQ 

the force Q is replaced by two components - and — , 

R R 

A A 

which form with it the angles (q, r) and (p, e). We 



NOTE. 207 



p2 q2 ^ 

can suppose that the forces — , — are directed along the 
same line as the resultant e, and then the two forces 

PQ 

equivalent to — will each form with the direction of R 

an angle equal to (p, q). Hence, they will form between 

A 
them an angle equal to double (p, q). Hence, since the 

A . . 

angle (p, q) is a right angle, by hypothesis, the forces 

PQ 
equivalent to — will be equal, but directly opposed. 

R 

Consequently, they will be in equilibrium ; and for the 
forces — , — , directed along the same line, we can sub- 

stitute only the forces P, Q, or their resultant R. Hence, 
we shall have the equation, 

p2 q2 

R=— + — , 
R R 

from which formula (1) is immediately deduced. 
This demonstration is due to Daniel Bernoulli. 



LEMMA III. 

The resultant R of the tivo forces P, Q, 2vhich intersect 
at right angles, is represented, not only in intensity, as 
we have proved above, but also in direction, by the diago- 
nal of the rectangle constructed upon the two components. 

Demonstration. This proposition is evident in the 
case where the forces p, q are equal to each other. 



208 STATICS. 

Then the resultant R should necessarily divide the angle 

A 
(p, q) into two equal parts, and we have, by virtue of 

Lemma II, 

R2=2p2, or R=pv/2. 

It is also easy to demonstrate. Lemma III, in the case 
where we suppose q^=2p^, or q=pn^2. Thus, we will 
consider three forces, equal to p, directed along three 
lines which are perpendicular to each other. These 
three forces will be represented by three sides of a cube 
which meet in the same summit. Moreover, the result- 
ant of two of these forces being equal to p^2, and di- 
rected along the diagonal of one of the forces of the 
cube, the resultant Q of the three forces will of necessity 
be included in the whole plane, which will contain one 
of the forces p and the diagonal of the square con- 
structed upon the other two. Now, there are three 
planes of this kind, and these three planes intersect in 
the diagonal of the cube. Hence, the resultant of the 
three forces p, or what is the same, the resultant of the 
forces P and pv^2, which intersect at right angles, will 
be directed along the diagonal of the cube, which is at 
the same time the diagonal of the rectangle constructed 
upon the forces p and p^^2. 

It might be proved, precisely in the same manner, 
that if we designate by m a whole number, and suppose 
Lemma III to be demonstrated in the case where we 
have Q=P>^m, the resultant of three forces, respect- 
ively equivalent to 

p, p, v^m^ 



NOTE. 209 

and represented by three lines perpendicular to each 
other, will be directed along the diagonal of the rect- 
angular parallelopipedon, -which will have these same 
lines for sides. From this we conclude, in the admitted 
hypothesis, that Lemma III will also subsist, if we take 
for Q the resultant of the forces P and v^m^ that is to 
say, if we make Q=P^??^^-l. Besides, Lemma III is 
evident, when we have Q=P, or, what is the same, m=l. 
Hence, this lemma will also subsist, if we take 
Q=:p\/l-fi=p\/2, or Q=p^^2+l=pv^3, etc., 

or, in general, Q=pv^w, m being any whole number. 

Let us conceive now, that m and n designate two 
whole numbers ; and construct a rectangular parallelo- 
pipedon, which has for its sides three lines representing 
the three forces 

p, v^rn^ v^n. 

The resultant of these three forces will evidently be 
contained; 1st, in the plane which includes the force 
v^n and the diagonal p^^m+1 of the rectangle con- 
structed upon the forces p, pv^m; 2d, in the plane 
which includes the force v^m and the diagonal v^n+1 

of the rectangle constructed upon the forces p, v^n. 
Hence, this resultant will be directed along the diagonal 
of the parallelopipedon ; and the plane, which contains 
the same resultant with the force p, will intersect the 
plane of the two forces P^^m, v^n along the diagonal 
of the rectangle constructed upon these two forces. 
Hence the resultant of the forces p^^m, pv^n, which 
evidently should be comprised in the plane in question, 

18* 



210 STATICS. 

will be directed along this last diagonal. Hence Lemma 
III will subsist, when we replace the forces P and Q 
by two forces equal to Q^^m, p^tz^, that is to say, by 
two forces whose squares are to each other in the ratio 
of m to n, Fonce, Lemma III will also subsist between 
the forces p and Q, if we suppose 

Q^ m \m 

-~=— or Q=pJ — 

'p^ n ^ n 

Now, let Q=P:?;, x designating any number whatever. 
We can vary the whole numbers m and t?-, in such a 

manner that the ratio — will converge to the limit x^^ 

n 

and it is evident, that, in this case, the resultant of the 






forces P and Ps/— j directed along two lines perpendicu- 

lar to each other, will tend more and more to coincide, 
in intensity and direction, on the one hand with the re- 
sultant of the forces p, vx^ and, on the other, with the 
diagonal of the rectangle constructed upon these two 
forces. Hence, the resultant of the forces P, vx will 
be represented by the diagonal in question. 



Corollary L 

If the force R be represented by the length ab laid 
off from its point of application (a) upon the line along 
which it acts, and if we draw through the point (a) two 
axes perpendicular to each other, we may substitute for 
the force R the two forces represented in intensity and 



NOTE. 211 



direction by the projections of the line ab upon the 
two axes. 



Corollary IL 

Let us conceive now, that two forces p, q, being applied 
to the same point (a), and represented by two lines AB, 
AC, which form any angle between them, to be traced in 
the plane of these two forces, two axes, one of which 
coincides with the diagonal of the parallelogram to 
which they serve as sides, and the other perpendicular 
to this diagonal. We can substitute for the two forces 
p, Q, the four forces represented in intensity and di- 
rection by the projections of the lines ab, ac upon the 
two axes. Now, of these four forces, two, being directly 
opposed, will be in equilibrium. The other two, directed 
along the diagonal of the parallelogram, will be added 
together, and will give, for their sum, a force represented 
in intensity and direction by this same diagonal. Hence 
we may enunciate the following proposition: 



THEOREM. 

The resultant R of two forces p, Q shnultaneously ap- 

^^ plied to a material point (a) and directed in any manner 

whatever^ is represented in intensity and direction hy 

the diagonal of the parallelogram constructed upon these 

two forces. 



212 STATICS. 



Corollary I. 



As the diagonal R of the parallelogram constructed 
upon the two forces p, Q is at the same time the third 
side of the triangle, which is formed by drawing through 
the extremity of the first force a line equal and parallel 
to the second, and as the angle in this triangle, opposite 

A 
to the side R, is the supplement of the angle (p, q), we 

have necessarily, by virtue of a known formula in 

trigonometry, 

A 
r2=^p2+q2+2pq cos (p, q) (2). 

Corollary II. 

In the case where the forces p, q become equal to 
each other, their resultant R is represented in intensity 
and direction by the diagonal of the lozenge constructed 
upon these same forces. Then the formula (2) re- 
duces to 

A 
r2=2pM1+cos(p, q)|. (3) 

A 
Moreover, by supposing (p, r)=^, we will find, in the 

A 
present case, (p, q)=^; and as we have, in general, 

cos 29=2 cos^ 0—1, (4) 

the equation (3) will give r=2p cos 0, or, what is the 
same, 

A 
r=2p cos (p, r). (5) 



NOTE. 213 

For the rest, we may be directly assured that the 
second member of the formula (5) represents the di- 
agonal of the lozenge constructed upon the two forces 
©qual to P. 

It is easy to demonstrate the theorem of page 211, 
for the case in which the forces P, Q have any ratio to 
each other, when we have once established this theorem 
for the case where we have Q=p, that is to say, when 
formula (5) is established. Now, we can give a direct 
demonstration of this formula, which, in fact, is deduced 
from equation (4), but which appears to merit special 
remark. I will explain it in a few words. 

Let us admit that formula (5) is verified for the case 

A 

where we have (p, r)— '^'j ^ designating either a right 

angle or an acute angle. I say that it will still subsist, 

if we suppose 

A ^ A ^ 
(p,r)=-, or(p,ii)=2- 

A 
Now, in these two hypotheses, the angle (p, q), included 

between the directions of the two equal forces P, Q, 

will be equivalent to one of the angles t*, 7t— t* ; and we 

can prove by reasoning as in Lemma II, that we can 

substitute in the system of the two forces P, Q, or for 

p2 

their resultant R, four components equal to — , among 

which, two will be directed along the same line and in 
the same direction as the force R, while the other two 
will each form with the resultant R an angle equivalent 

A 
to (p, r), that is to say, to t or to Tt—t* Now, since we 



't 



214 STATICS. 

suppose formula (5) to be verified in the case where we 

A 
have (p, r)='^5 the last two components evidently can 

be replaced by a single force equal to 2— cos t-j and 

situated in the direction of the resultant R, or in the 
opposite direction. Consequently, we will find defini- 
tively, 

p2 p2 p2 

R=2— =b2— cos 1^=2— (l±cos t) ; 

H R R 

or, 

R2=2p2 (ld=cos *), (6) 

the sign ± being necessarily reduced to the sign+, in 

A ^ 
the case where (p, r)=9j and to the sign — in the case 

where (p, e)=-— 5. Since we will also get the formula 
(4) by placing successively 

1+cos t=^2 cos^ 5, l— cos Tf=2 cos^ --^, 
equation (6) will give in the first case 

R=2 COS 05 

and, in the second, 

R = COS — - — 

A 

Hence, if equation (4) subsists when we attribute to the 
angle (p, r) the value t*, it will also subsist when we 



NOTE. 215 

attribute to the same angle one of the values ^5 ^7— • 

Now, this equation is verified when we suppose (p, e,)=5, 

since we have evidently in this hypothesis ^=0, and 

A 
cos (p, r) =0. Hence, it will be equally true if we 

suppose 

(p, R)=2. 2=4' 
and, consequently, if we take 

Hence it will also be true, if we attribute to the angle 

A 
(p, r) one of the values, 

1 ii 7i 1 S^t 37t \, %Tt. 5rt 

2'8""i6' 2' 8"~i6' 2''''~8"^~16' 

1 , _ ?i;,_7?t 

2 *'''~i6^~i6' ■ ' ' 

By continuing in the same manner, it may be proved 

A 
that formula (5) generally has place when the angle (p, r) 

receives a value of the form — - — , n representing any 

A 

whole number, and 2m+l an odd number less than 2^. 

If we now represent by an acute angle taken at will, 

we can vary the whole numbers m and n in such a man- 

2m+l 
ner that the ratio — ^ — will indefinitely approach the 



216 STATICS. 

limit e ; and the resultant r will tend more and more 
to coincide, on the one hand, with a force equivalent to 
2p cos e^ and, on the other hand, with the resultant of 
two forces equal to p, which would form between them 
an angle double of 0. Hence, this last resultant will 
be represented in magnitude by 2p cos 6?, and will also 
verify formula (5). 



THE END. 



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